C語言及實驗作業(yè)參考做法

C語言及實驗作業(yè)參考做法順序結(jié)構(gòu)程序設計3.04個字符描述用4個變量表示4個字符,從鍵盤上輸入4個不同的小寫字符,輸出它們對應的大寫字符。輸入數(shù)據(jù)4個小寫字符輸出數(shù)據(jù)4個大寫字符輸入示例abed（中間沒有空格）輸出示例ABCD3.1平均成績描述已知某個同學的語文、數(shù)學、英語成績,計算該同學的平均分輸入數(shù)據(jù)3個整數(shù)輸出數(shù)據(jù)1個實數(shù),精確到小數(shù)點后2位輸入示例806792輸出示例79.673.2找錢描述輸入顧客在超市.消費金額和顧客支付金額,輸出應找的錢數(shù)。輸入數(shù)據(jù)兩個實數(shù)輸出數(shù)據(jù)1個實數(shù),精確到小數(shù)點后2位輸入示例89.67100輸出示例10.333.3正弦值描述輸入角度a,輸出sina輸入數(shù)據(jù)1個表示角度的整數(shù)輸出數(shù)據(jù)!個實數(shù),精確到小數(shù)點后2位輸入示例35輸出示例0.57分支程序設計上機作業(yè):

描述4.I計算方程ax2+bx+c=0的實數(shù)根輸入三個實數(shù)a,b,c由其組成一個方程ax2+bx+c=0?求該方程的實數(shù)根。若方程有實數(shù)根,輸出這兩個根,其中,X1為數(shù)值小的根,x2為數(shù)值大的根;若沒有實數(shù)根,則輸出"norealnumberroots"輸入數(shù)據(jù)三個用逗號格開的實數(shù)輸出數(shù)據(jù)x1=小根,x2=大根,取小數(shù)點后兩位輸入示例1.0,-6.0,8.0輸出示例x1=2.00,x2=4.00輸入示例3.0,1.0,1.0輸出示例norealnumberroots1,2,-3x1=-3.00,x2=1.001,21x1=1.00,x2=1.002,3,2norealnumberroots1,-5.5,7x1=2.00,x2=3.50標題4.3三角形判斷描述輸入三條邊的長，輸出這三條邊是否構(gòu)成等邊三角形、等腰三角形、直角三角形、普通三角形,還是不能構(gòu)成三角形（按最嚴格條件輸出，如輸出等邊三角形不輸出等腰二角形、普通三角形）輸入數(shù)據(jù)整數(shù)a,b,c的值輸出數(shù)據(jù)等邊三角形、等腰三角形、直角三角形、普通三角形，次為1、2、3、4、5或不能構(gòu)成三角形依輸入示例555輸出示例1^include<stdio.h>intmain(){inta,b,c;scanf("%d%d%d”,&a,&b,&c);if(a==b&&b==c)printf("1");elseif(a+b>c&&a+c>b&&b+c>a)if(a==b||b==c||a=c)printf("2");elseif(a*a+b*b=c*c||b*b+c*c=a*a||a*a+c*c=b*b)printf("3");elseprintf1'4);elseprintf15);return0;)測試數(shù)據(jù)

33226810335741285101010166125標題4.4以12小時制輸出時間描述輸入24小時制時間,輸出12小時制時間。十二小時制是一個時間規(guī)則把?日二十四小時分為兩個時段,分別為上午（AM）和F午（PM）。每個時段由十二個小時構(gòu)成,以數(shù)字12、1、2、3、4、5、6、7、8、9、10、11依次序表示。上午時段由午夜至中午,而下午時段由中午至午夜。二十四小時制（的規(guī)則），是把毎II由午夜至午夜共分為二十四個小時,從數(shù)字0至23。輸入數(shù)據(jù)整數(shù)：整數(shù)（24小時制時間）輸出數(shù)據(jù)整數(shù):整數(shù)（12小時制時間）輸入示例18:20輸出示例6:20PM測試數(shù)據(jù)12:2012:20PM13:201:20PM18:206:20PM0:2012:20AM7:207:20AM12:0012:00PMintmain(){inthour,minute;scanf(4€%d:%d,,,&hour,&mmute);if(hour>12&&hour<24)printR"%d:%dpm\n",hour-12,minute);elseifi(hour>0&&hour<12)printfC*%d:%dam也”,hour,minute);elseifi(hour=0)printR"％d:%dam\n",hour+l2,minute);elseifi(hour==12)printf(*t%d:%dpm\n",hour,minute);elseprintff'inputerror\n")；return0;)

標題4.5計算函數(shù)的值描述當X<1時,y=X當l<x<10時,y=2xT當x>10時,y=3xTl輸入數(shù)據(jù)整數(shù)X的值輸出數(shù)據(jù)y=ー個整數(shù)表示函數(shù)的值輸入示例5輸出示例y=9#include<stdio.h>intmain(){intx,y;scanf("%d",&x);if(x<l)y=x;elseif(x<10)y=2*x-l;elsey=3*xTl;printf(*%d*,y);return0;)測試數(shù)據(jù)-5y=-510y=196y=ll15y=341y=l標題4.6數(shù)的分離描述給ー個不多于3位的正整數(shù)(1)計算它是兒位數(shù)(2)分別打印出每一位數(shù)字(3)逆序打印出各位數(shù)字輸入數(shù)據(jù)?個不多于3位的正整數(shù)輸出數(shù)據(jù)位數(shù)從高位到低位的各位數(shù)字從低位到高位的各位數(shù)字輸入示例456輸出示例3456654提示123/100=?

23/10=?123%10=?據(jù)據(jù)據(jù)數(shù)數(shù)試試試6783678876據(jù)據(jù)據(jù)數(shù)數(shù)試試試45245548188#include<stdio.h>intmain(){intp,t,s;inti,bit;//printf("inputanumber(0-999):");scanf("%d",&p);gewei=p%10; 〃個位shiwei=p/10%10; 〃十位baiwei=p/100;〃百位if(baiwei)〃如果是3位數(shù){printf(z，%d\n",3);printf("%d%d%d\n",baiwei,shiwei,gewei);printf("%d%d%d\n",gewei,shiwei,baiwei);)elseif(shiwei)〃如果是2位數(shù){printf("%d\n",2);printf("%d%d\n",shiwei,gewei);printf("%d%d\n",gewei,shiwei);)elseprintf("%d\n%d\n%d\n",1,gewei,gewei):〃是1位數(shù)return0;1測試數(shù)據(jù)56 25623465234標題4.7輸出等級分（用if語言實現(xiàn)）描述輸入一個百分制成績,輸出對應的等級分90-100為A80-89為B70-79為C60-69為D0-59為E3234432

輸入數(shù)據(jù)ー個整數(shù)輸出數(shù)據(jù)ー個大寫字符輸入示例95輸出示例A測試數(shù)據(jù)100A89B70C62D0E59E90A93A80B標題4.8計算建筑高度描述有4個圓塔,圓心分別為(2,2),(-2,2),(-2,-2),(2,-2),圓半徑為10這4個塔的高度為10m,塔以外無建筑物。輸入任一點的坐標,求該點的建筑高度(塔外高度為0)。輸入數(shù)據(jù)以(X,y)形式作為點的坐標輸入輸出數(shù)據(jù)輸出高度輸入示例(3,1)輸出示例10提示注意輸入坐標的格式,括號為半角符號輸入:(0.5,0.7)輸出:〇輸入：(2.1,2.3)輸出：10輸入:(-2,-1) 輸出:10輸入：(-3,2) 輸出：10^include<stdio.h>main(){intxl=2,yl=2;floatx,y;//printfC^inputthepositionx&y:");scanfC(%f,%f)",&x,&y);x=fabs(x);y=fabs(y);if((x-xl)*(x-xl)+(y-yl)*(y-yl)<=l)printf("%d\n",10);elseprintf("0\n");

return0;)測試數(shù)據(jù)(1.6J.7)10(-1.0,2.0)10(2.3,-2.6)10(-2.2,-1.3)10(-1.0,-1.0)0(7.0,8.0)0(3.0,2.0)1()標題4.9輸出等級分（用switch語言實現(xiàn)）描述輸入一個百分制成績,輸出對應的等級分90-100為A80-89為B70-79為C60-69為D0-59為E輸入數(shù)據(jù)?個整數(shù)輸出數(shù)據(jù)ー個大寫字符輸入示例86輸出示例Bftinclude<stdio.h>intmain(){intscore;chargrade;printf<inputascore:*)：scanf(*%d*,&score);switch(score/10){case10:case9:grade='A';break;case8:grade='B';break;case7:grade='C';break;case6:grade='D';break;case5:case4:case3:case2:case1:case0:grade='E'：break：Jprintf(*hisscoreclassis%c\n*,grade)：

return0;標題4.10格式輸出時間描述以mm/dd/yy格式輸入年月日,輸出:Datedthis dayof_,20_〇輸入數(shù)據(jù)mm/dd/yy輸出數(shù)據(jù)Datedthis dayof ,20_輸入示例7/19/12輸出示例Datedthis19thdayofJuly,2012測試數(shù)據(jù)3/1/12Datedthis1stdayofMarch,20124/2/26Datedthis2nddayofApril,20269/7/8Datedthis7thdayofSeptember,20085/3/14Datedthis3rddayofMay,20141/30/06Datedthis30thdayofJanuary,20066/23/56Datedthis23rddayofJune,205612/21/10Datedthis21stdayofDecember,201010/31/4Datedthis31stdayofOctober,20048/22/09Datedthis22nddayofAugust,2009#include<stdio.h>intmain()(intmonth,day,year;printf(HEnterdatemm/dd/yy”)；scanf(n%d/%d/%dn,&month,&day,&year);printf(HDatedthis%dH,day);switch(day)(case1:case31:printf(nstn);break;case2:printff'nd")；break;case3:printff'rd")；break;default:printf(nthM);break;)printf(Mdayofn);switch(month)casel:printfi(nJanuaryH);break;case2:printf(HFebruaryH);break;case3:printf(nMarchM);break;case4:printf(HAprir,);break;case5:printf(nMayn);break;case6:printf{nJunen);break;case7:printf(nJulyH);break;case8:printf{”August”)；break;case9:printfi(nSeptemberH);break;case10:printf(nOctobern);break;case11:printfi[,,Novembern);break;case12:printf(,,Decembern);break;)printf(",20%02d.\n",year); //printf(",20%.2d.\n",year);return0;循環(huán)程序設計標題5.2統(tǒng)計一行字符中各類字符的個數(shù)描述輸入一行字符,統(tǒng)計其中英文字母、空格、數(shù)字和其他字符的個數(shù)輸入數(shù)據(jù)一行以回車結(jié)束的字符輸出數(shù)據(jù)四個分別表示英文字母、空格、數(shù)字和其他字符的個數(shù)的整數(shù)輸入示例123abcd****ABCD輸出示例8234ttinclude"stdio.h"main(){intnum0=0,numa=0,num=0,other=0;charch;//num0:0-9;numa:alpha;num:space;scanf("%c”,&ch);while(ch!=,\n){if(ch>='O'&&chくニ’9')num0++;elseif(ch>='a'&&ch<='z'||ch>='A'&&ch<='Z')numa++;elseif(ch='')num++;elseother++;scanf("%c”,&ch);)printf("%d%d%d%d”,numa,num,numO,other);return0：

)測試數(shù)據(jù)hhhh8888****43446666KKK***kkk6043****hhhh4804*&K9H8&2623?234 ****04441234ABCD4040標題5.3計算aaaaa描述計算S(n)=a+aa+aaa+ 其中a是ー個數(shù)字輸入數(shù)據(jù)兩個分別表示a和n的整數(shù)輸出數(shù)據(jù)ー個表示S(n)的整數(shù)輸入示例35輸出示例37035提示2*10+2=?22*10+2=?升includeくstdio.h>intmain(){inta,n,i;longal1=0,t;scanf("%d%d*,&a,&n);t=a;for(i=l;i<=n;i++){all=all+t;t=t*10+a;)printfall);return0;)測試數(shù)據(jù)36370368400637389711111103060標題5.6找出1000以內(nèi)的所有完數(shù)描述ー個數(shù)等于它的所有因子之和,這個數(shù)就稱為"完數(shù)'找出1000以內(nèi)的所有完數(shù)

輸入數(shù)據(jù)無輸出數(shù)據(jù)完數(shù)及它的所有因子（用逗號格開）輸入示例無輸出示例6itsfactorare1,2,328itsfactorare1,2,4,7,14496itsfactorare1,2,4,8,16,31,62,124,248#includestdio.h"main(){inta,b,k,s;for(k=2;k<1000;k++){〃判斷k是否是完數(shù)s=0;for(a=1;a<=k/2;a++)if(k%a=0)s+=a! 〃計算k的因子之和if(s=k) 〃k是完數(shù){printf(*%ditsfactorsare1,k);for(a=2;a<=k/2;a++)if(k%a=0)printf(",%d",a);〃輸出所有因子printf(*\n*);)})標題5.7計算分數(shù)序列前n項之和描述計算分數(shù)序列2/1,3/2,5/3,8/5,13/8,21/13, 前n項之和輸入數(shù)據(jù)整數(shù)n輸出數(shù)據(jù)ー個實數(shù),精確到小數(shù)點后第3位輸入示例20輸出示例32.660測試數(shù)據(jù)58.392813.2441524.570intmain(){floatfenzi=2,fenmu=l,temp;intn,i;doublesumニ〇;scanf("%d",&n);for(i=0;i<n;i++)

{ sum+=fenzi/fenmu;printf("%f\n〃,fenzi/fenmu);temp=fenzi;fenzi=fenzi+fenmu;fenmu=temp;)printfC*%.3f”,sum);return0;標題5.8猴子吃桃問題描述猴子第一天摘下若干個桃子,當即吃了一半,還不過癮,又多吃了一?個.第二天早上又將剩下的桃子吃掉一半,又多吃了一個.以后每天早上都吃了前一天剩下的一半零一個.到第n天早上想再吃時,就只剩下ー個桃子了.問第一天共摘了多少桃子.輸入數(shù)據(jù)整數(shù)n輸出數(shù)據(jù)表示第一天桃子個數(shù)的整數(shù)輸入示例10輸出示例1534#include<stdio.h>main(){intday,all,n;scanf("%d",&n);all=l;for(day=n-l;day>0;day-)al1=2*(al1+1);//printf(*thefirstnumberis);printf("%d”,all);}測試數(shù)據(jù)10153420157286242211標題"9與n無關(guān)數(shù)的平方和描述?個正整數(shù),如果它能被n整除,或者它的十進制表示法中某個數(shù)位上的數(shù)字為n,則稱其為與n相關(guān)的數(shù)。現(xiàn)求所有小于等于m(m<100)的與n無關(guān)的正整數(shù)的平方和。

輸入數(shù)據(jù)輸入為一正整數(shù)m,(m<100)和個位整數(shù)n輸出數(shù)據(jù)輸出小于等于m的與n無關(guān)的正整數(shù)的平方和輸入示例217輸出示例2336測試用例56947621243236635103524605ttinclude<stdio.h>intmain()(intm,n;intsum=O;inti,ge,shi;scanf("%d%d”,&m,&n);for(i=l;i<=m;i++){ge=i%10;shi=i/10;if(ge==n||shi=n||i%n==O)continue;sum+=i*i;}printfsum);return0;)標題5.10輸出菱形圖案描述輸出圖案輸入數(shù)據(jù)無輸出數(shù)據(jù)菱形圖案輸入示例無輸出示例********提示找規(guī)律上下部分規(guī)律不同

標題5.11輸出菱形圖案2選做?描述輸入ー個整數(shù)n和一?個字符c,輸出菱形圖案.菱形圖案由字符c構(gòu)成,圖案總共2n+l行輸入數(shù)據(jù)ー個整數(shù)和一個字符輸出數(shù)據(jù)菱形圖案輸入示例3*輸出示例********提示上下n行對稱#include<stdio.h>intmain(){intk,i,j,n;charc;scanf("%d%c”,&n,&c);for(i=l;i<=n+l;i++){for(j=l;j<=n+l-i;j++)putcharf*);for(j=l;j<=2*i-l;j++)putchar(c);printf(*\n*);)for(i=l;i<=n;i++){for(j=l;j<=i;j++)putchar(*');for(j=l;j<=(n+l-i)*2-l;j++)putchar(c);putchar('\n);)return0;)測試數(shù)據(jù)2********?5$$$$$

①①①史ぐぐぐ①①①ぐぐぐくぐ①ぐぐぐぐummm〇〇ふ〇ccccccc》》》》》》》》》》》》$$$$6JJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJ標題 !5.14九九乘法表描述輸出上三角形式的九九乘法表輸入數(shù)據(jù)無輸出數(shù)據(jù)1234567894 6 8 10 12 14 16 189 12 15 18 21 24 2716 20 24 28 32 3625 30 35 40 4536424854495663647281intmain(){introw,col;for(row=l;row<10;row++){for(col=l;col<row;col++)printf("%4c",'');for(col=row;col<10;col-H-)printf("%4d”,row*col);printfC\nO;)

return0;)標題5.12求最大描述輸入ー個正整數(shù)repeat(0くrepeatく10),做repeat次下列運算:輸入ー個正整數(shù)n,再輸入n個整數(shù),輸出最大值。輸入數(shù)據(jù)正整數(shù)repeat,輸入ー個正整數(shù)n,再輸入n個整數(shù)輸出數(shù)據(jù)最大值輸入示例例:括號內(nèi)是說明輸入3(repeat=3)42123-1000491-183135輸出示例12395標題5.13與7無關(guān)數(shù)的平方和描述?個正整數(shù),如果它能被7整除,或者它的十進制表示法中某個位數(shù)上的數(shù)字為7.則稱其為與7相關(guān)的數(shù).現(xiàn)求所有小于等于n(n<100)的與7無關(guān)的正整數(shù)的平方和.輸入數(shù)據(jù)輸入為一正整數(shù)n,(n<100)輸出數(shù)據(jù)輸出小于等于n的與7無關(guān)的正整數(shù)的平方和輸入示例21輸出示例2336Hinclude<stdio.h>intmain()(intn;intsum=0;inti,ge,shi;scanf("%d",&n);for(i=l;i<=n;i++){ge=i%10;shi=i/10;if(ge==71|shi=71|i%7=0)continue;sum+=i*i；)printfsum);return0;)惻試數(shù)據(jù)2123367911377078855965（）書面作業(yè):1.從鍵盤上輸入10個整數(shù),計算這些整數(shù)的和?水仙花數(shù)。ー個3位數(shù),若其各位數(shù)字之和等于它本身,這個數(shù)就稱為水仙花數(shù).輸出所有的水仙花數(shù)。(如153)#includeくstdio.h>intmain()〃方法一{intnum,bai,shi,ge;for(num=100;num<1000;num++){bai=num/100;shi=num/10%10;ge=num%10;if(num==bai*bai*bai+shi*shi*shi+ge*ge*ge)printf("%d\n",num);}return0;)〃方法二intmain(){inti,j,k;for(i=l;i<10;i++)for(j=0;j<10;j++)for(k=0：k<10;k++)if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)printf(*%d\n*?100*i+j*10+k);return0;)■計算分數(shù)序列2/1,3/2,5/3,8/5,13/8,21/13, 前20項之和。#include<stdio.h>intmain()floatsum,fenzi,fenmu,a;inti;fenzi=2;fenmu=l;sum=0;for(i=l;i<=20;i++)(sum+=fenzi/fenmu;a=fenmu;fenmu=fenzi;fenzi+=a;}printf("%.3f*,sum);return0;}■迭代法求平方根。利用公式x(n+l)=l/2*(x(n)+a/x(n))計算a的平方根.要求|x(n+l)-x(n)|<le-5ttinclude<stdio.h>ttinclude<math.h>mainO{floatxO,xl,a;printf(*inputa=")；scanf("%f",&a);xl=a/2;do{x0=xl;xl=(x0+a/x0)/2;}while(fabs(xl-x0)>le-5);printf("thesqrtof%fis%f\n",a,xl);).百錢買百雞:雞翁ー,值錢五;雞母ー,值錢三;雞雛三,值錢ー,百錢買百雞,問翁、母、雛各兒何?.已知某年一月一日是星期幾,按照SUNMONTUSWEDTHUFRISTA的格式輸出該年某個月的月歷。(指定?個月的天數(shù)和這個月?日是星期兒,按照SUNMONTUSWEDTHUFRISTA的格式輸出這個月的月歷。)#include<stdio.h>intmain(){intyear,month,week;intleap,i;intday,firstweek,monthday;printf(z，Enterayear:z，);scanf(〃％d〃,&year);〃輸入年份printf("Enterweekonfirstday:");scanf(級d〃,&week);〃輸入該年1月1日星期幾printf("Whichmonth:");scanf("%d",&month); 〃輸入月份if(year%4==0&&year%100!=:0|year%400==0)leap=l;elseleap=0;〃判斷該年是否是閏年day=0; 〃計算年year中月month前有多少天for(i=l;i<month;i++)switch(i){case1:case3:case5:case7:case8:case10:case12:day+=31;break;case2:if(leap)day+=29;elseday+=28;break;case4:case6:case9:day+=30;break;)firstweek=(week+day)%7;〃計算該月第一天星期兒switch(month) 〃計算月month有多少天{case1:case3:case5:case7:case8:case10:monthday=31;break;case2:if(leap)monthday=29;elsemonthday=28;break;case4:case6:case9:ase11:monthday=30;break;)printfCSUNMONTUSWEDTHUFRISTA\n〃);〃輸出月歷標題for(i=0;i<firstweek;i++) 〃輸出第一天前的空格printf("%4c",'');for(i=l;i<=monthday;i++) 〃輸出該月的每一天{printf(，，%4d/，,i);firstweek=(firstweek+l)%7;if(firstweek=0&&i!二monthday)printf〈'n");}printf(A'\n〃)；return0;).輸入兩個分數(shù),計算它們的和,并以最簡分式的方式輸出intmain(){intfenzil,fenmul,fenzi2,fenmu2,fenzi,fenmu;printf(〃請輸入第一個分數(shù)的分子分母(格式:分子/分母)：");scanf("%d/%d",&fenzil,&fenmul);printf(〃請輸入第二個分數(shù)的分子分母(格式:分子/分母):〃);scanf(〃％d/%d〃,&fenzi2,&fenmu2);fenzi=fenziI*fenmu2+fenzi2*fenmul;fenmu=fenmu1*fenmu2;if(fenzi%fenmu==0)printf("和為％d”,fenzi/fenmu);else{inti,min;min=(fenzi<fenmu)?fenzi:fenmu;for(i=min;i>l;i--)if(fenzi%i==0&&fenmu%i==0)break;fenzi/=i;fenmu/=i;printf(〃和為％d/%d\n〃,fenzi,fenmu);return0;

intmain()intfenzi1,fenmul,fenzi2,fenmu2,fenzi,fenmu;printf（"請輸入第一個分數(shù)的分子分母（格式:分子/分母）:〃）;scanf &fenzi1,&fenmul);printf（"請輸入第二個分數(shù)的分子分母（格式:分子/分母）:〃）;scanf(〃％d/%d〃,&fenzi2,&fenmu2);fenzi=fenzil*fenmu2+fenzi2*fenmul;fenmu=fenmul*fenmu2;if(fenzi%fenmu==O)printf(〃和為%d”,fenzi/fenmu);else{inti;for(i=2;i<=fenzi&&i<=fenmu;)if(fenzi%i==O&&fenmu%i==O)fenzi/=i,fenmu/=i;elsei++;printf("和為%レ％(1ヽぺ［fenzi,fenmu);)return0;.輸入若干個日期（以〇/〇/〇結(jié)束輸入）,輸出最早的日期.ー個エ廠制造的產(chǎn)品形狀都是長方體,它們的高度都是h,長和寬都相等,ー共有六個型號,他們的長寬分別為ド1,2*2,3*3,4*4,5*5,6*6。這些產(chǎn)品通常使用ー個6*6*h的長方體包裹包裝然后郵寄給客戶。因為郵費很貴,所以工廠要想方設法的減小每個訂單運送時的包裹數(shù)量。他們很需要有一個好的程序幫他們解決這個問題從而節(jié)省費用?，F(xiàn)在這個程序由你來設計。輸入輸入數(shù)據(jù)包括幾行,每一行代表?ー個訂單。每個訂單里的一行包括六個整數(shù),中間用空格隔開，分別為け1至6*6這六種產(chǎn)品的數(shù)量。輸入文件將以6個〇組成的一行結(jié)尾。輸出除了輸入的最后一行6個〇以外,輸入文件里每一行對應著輸出文件的一行,每一行輸出ー個整數(shù)代表對應的訂單所需的最小包裹數(shù)。樣例輸入樣例輸出數(shù)組上機作業(yè):

標題6.0相鄰兩個數(shù)比較描述輸入8個整數(shù),從前往后比較相鄰的兩個整數(shù),如果前面的數(shù)大,則交換。輸出最后這8個數(shù)的值。要求用數(shù)組實現(xiàn)。輸入數(shù)據(jù)8個整數(shù)輸出數(shù)據(jù)比較并調(diào)整位置后的8個整數(shù)輸入示例87654321輸出示例76543218標題6.1最小數(shù)的位置描述輸入10個整數(shù),輸出從第n個數(shù)到最后一個數(shù)中最小數(shù)的位置（下標）,要求用數(shù)組實現(xiàn)。輸入數(shù)據(jù)10個整數(shù)及n值輸出數(shù)據(jù)最小數(shù)的位置（0-9）輸入示例8715432922333輸出示例6標題6.2選擇排序描述用選擇法對!0個字符排序輸入數(shù)據(jù)10個字符輸出數(shù)據(jù)排好序的10個字符輸入示例fgthyjpdsa輸出示例adfghjpsty#includeくstdio.h>intmain(){chara[10];inti,j,k,t;for(i=0;i<=9;i++)scanf &a[i]);for(j=0;j<=8;j++)ik=j;for(i=j+l;i<=9;i++)tif(a[i]<a[k])k=i;{t=a[j];a[j]=a[k];a[k]=t;}))for(i=0;i<=9;i++)printf("%c",a[i]);

return0;)測試數(shù)據(jù)poiuytrewqeiopqrtuwyyuiopqwerteiopqrtuwyewqrtyuiopeiopqrtuwy標題6.3字符串排序描述輸入10個字符串,按從小到大的順序排序后輸出輸入數(shù)據(jù)10個字符串輸出數(shù)據(jù)從小到大有序的10個字符串輸入示例qweasdzxchgupcctvarenotgotohead輸出示例areasdcctvgotoheadhgnotqweupzxcM1一.提示要求用冒泡排序方法^defineN10intmain(){inti,j,k；chara[N][20],t[20];printf(uinput10characters:n);for(i=0;i<N;i++)gets(a[i]);for(i=0;i<N-l;i++)〃做N-1趟for(j=0;j<N-l-i;j++)〃j表示比較范圍if(strcmp(a[j],a[j+l])>0)〃前面的大,后面的小就交換{strcpy(t,a[j]);strcpy(a[j],a[j+l]);strcpy(a[j+l],t);}

)//printf(athesortedstrings;for(i=0i<N;i++)puts(a[i]);return0;)測試數(shù)據(jù)qweareasdasdzxccctvhggotoupheadcctvhgarenotnotqwegotoupheadzxcnotareqwetyuiuoasdupcctvzxcgotoareheadasdhgcctvnotgotoqwetyuiuoheaduphgzxcareareheadasdhgcctvnotgotoasdheadcctvhggotonotupqwetyuiuozxcupqwetyuiuozxc90817263544536

271809標題6.4有序插入選做題?描述輸入ー個有9個元素的有序數(shù)組,再輸入一個數(shù),把這個數(shù)插入到數(shù)組中,插入后仍然有序輸入數(shù)據(jù)10個整數(shù),前9個數(shù)從小到大有序輸出數(shù)據(jù)10個有序整數(shù)（用%3d輸出）輸入示例246810121416187輸出示例246781012141618提示注意數(shù)組大小試驗以下3組輸入:24 6 8 10 12 14 16 18 724 6 8 10 12 14 16 18 124 6 8 10 12 14 16 18 27#include<stdio.h>intmainO{inta[10];inti,j,t;for(i=0;i<9;i++)scanf("%d",a+i);〃輸入9個數(shù)scanfC%d\&t);〃輸入要插入的數(shù)for(i=8;i>=0;i-)〃從后往前找插入位置,并把大的數(shù)往后挪if(a[i]<t)break;elsea[i+l]=a[i];a[i+l]=tJ〃插入for(i=0;i<10;i++)〃輸出插入后且有序的數(shù)printf(*%3d*?a[i]);return0;)測試數(shù)據(jù)35791113151719-2-23579111315171935791113151719835789111315171935791113151719893579111315171989357911131517199357991113151719標題6.5麥森數(shù)選做題?描述形如2ビ1的素數(shù)稱為麥森數(shù),這時P■?定也是個素數(shù)。但反過來不一定,即如果P是個素數(shù)。2匸1不一定也是素數(shù)。輸入P（1000くPく310000〇）,計算2匸1的位數(shù)和最后500位數(shù)字（用十進制

高精度數(shù)表示）輸入數(shù)據(jù)?個整數(shù)（1000到3100000之間）輸出數(shù)據(jù)?個表示為數(shù)的整數(shù)和500位數(shù)字輸入示例1279輸出示例38600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010407932194664399081925240327364085538615262247266704805319112350403608059673360298012239441732324184842421613954281007791383566248323464908139906605677320762924129509389220345773183349661583550472959420547689811211693677147548478866962501384438260291732348885311160828538416585028255604666224831890918801847068222203140521026698435488732958028878050869736186900714720710555703168729087〃方法一#include<stdio.h>intmain(){longp,i,j；intweishu,jinwei,temp;inta[501]={l};intb[10]={l,2,4,8,16,32,64,128,256,512}：scanf("%d",&p);weishu=(int)(p*logl0(2))+l;printf(〃%d\n,weishu);for(i=p;i>0;i-){jinwei=0;for(j=0：j<501;j++){temp=a[j]*2+jinwei;a[j]=temp%10;jinwei=temp/10;))a[0]=a[0]-l;for(i=499;i>=0;i—){printf("%d",a[i]);if(i%50=0)printf("\n");}return0：)〃方法二#includeくstdio.h>intmain()

{longp,i,j；intweishu,jinwei,temp,tempwei,maxwei;inta[501]={6,7,5,8,4,0,1);//pow(2,20)intb[20]={l,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288};scanf(*%d*,&p);weishu=(int)(p*0.301029996)+1;printf("%d\n”,weishu);tempwei=4;for(i=p-20;i>=20;i-=20){jinwei=0;maxwei=tempwei+7;for(j=0;jくmaxwei;j++){temp=a[j]*1048576+jinwei;a[j]=temp%10;jinwei=temp/10;)tempwei=maxwei;while(!a[tempwei])tempwei—;tempwei++;)jinwei=0;for(j=0;j<501;j++){temp=a[j]*b[i]+jinwei;a[j]=temp%10;jinwei=temp/10;)a[0]=a[0]-l;for(i=499;i>=0;i—){printfa[i]);if(i%50=0)printf(*\n*);)return0;)測試數(shù)據(jù)1279386000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000104079321946643990819252403273640855386152622472667048053191123504036080596733602980122394417323241848424216139542810077913835662483234649081399066056773207629241295093892203457731833496

6158355047295942054768981121169367714754847886696250138443826029173234888531116082853841658502825560466622483189091880184706822220314052102669843548873295802887805086973618690071472071055570316872908723456706175091758749604823898061457688134908498481443379845244090909456487305616671101689989456098921997639762390056969279502998524342367440385625520904619998576236759533232262662329986769573290230960153554633695576996080169509863407587570341140986605581804439936972360010686479698419780140852917244016307442988015223710746219079066565764352449263612274735610712331157103771392555942242222219919312707002186721962230074373985949034213210770600298709244867596250146874460672696764411321917278091515769112231935234567870612003994340612599267325001374753485605064000616257870563191365761754037739729971447589092377265589454831204799417501351148007478970832421166682773628245403011597738271514201843943775714341562094060948821182555685897587666737087144942508029281229181523868239158321596916626846590276831997315228590762640086991869925697383639278551928902017495894347433916858277863715119802281512077343402903585691063891149489242503197394628580327152327453066139700240916036810862096283053033270359418192652133262419820543標題6.6打印楊輝三角形描述輸出n行楊輝三角形輸入數(shù)據(jù)ー個整數(shù)n輸出數(shù)據(jù)n行楊輝三角形〇輸入示例8輸出示例111121133114641151010511615201561172135352171提示注意輸出10行

用為4d輸出每個數(shù)#include<stdio.h>intmainO{inta[14][14]={0};inti,j,n;scanfC%d”,&n);for(i=0;i<n;i++)a[i][0]=l,a[i][i]=l;for(i=2;i<n;i++)for(j=l;j<i;j++)a[i][j]=a[i-l][j-l]+a[i-l][j];for(i=0;i<n;i++){for(j=0;j<=i;j++)printf(*%4d*.a[i][j]);printf(*\n*);}return0;)測試數(shù)據(jù)101111 2 113 3 114 6 4 11 5 10 10 5 11 6 15 20 15 6 11 7 21 35 35 21 7 11 8 28 56 70 56 28 8 11 936841261268436 9 1標題6.7大整數(shù)的因子 選做題?描述已知正整數(shù)k滿足2<=k<=9,現(xiàn)給出長度最大為30位的十進制非負整數(shù)C,求所有能整除c的k.輸入數(shù)據(jù)若干個非負整數(shù)c,c的位數(shù)く=30每行一個c,當c=T時中止（不要對T進行計算!）輸出數(shù)據(jù)每ー個c的結(jié)果占一行1）若存在滿足c%k==0的k,輸出所有這樣的k,中間用空格隔開,最后ー個k后面有空格。2）若沒有這樣的k則輸出"none"輸入示例307213 -12356輸出示例 234689none提示 [ー個整型變量是放不下30位整數(shù)的#includeくstdio.h>ttdefineN30+1intgetlargenumber(intA[N])〃讀入一個大整數(shù),放在數(shù)組A中{charc,temp[N];inti,j,k,start,len;for(i=0;i<N;i++)A[i]=0;i=0：c=getchar();if(c=-)return0;do{temp[i++]=c;}while((c=getchar())!='\n');start=N-i;i一；for(j=N-l;j>=start;j—,i一)A[j]=temp[i]-f0';return1;)intis7bei(intA[N]) 〃判斷是否能被7整除〃方法是每一位不斷的減7,直到小于7為止〃如865,與165具有相同的特性(最高位ー7),165與25具有相同的特性(1*10+6-7-7=2),25與4具有相同的特性(2*10+5-7-7-7){inti,k;for(i=0：;i++)if(A[i])break; 〃找最高位while(i<N){k=A[i-l]*10+A[i];while(k>=7)k-=7; 〃這位數(shù)字只要超過7就不斷的減7A[i-l]=0;A[i]=k;i++；)i一;if(A[i]%7=0)return1;return0;)intmain(){inta[N];intbei[10]={0}J〃存放大整數(shù)是否為7的倍數(shù),如bei[7]=1表示能被7整除intsum,i,t,yes;

t=getlargenumber(a);〃讀取大整數(shù)while(t){memset(bei,0,sizeof(bei));〃將bei的值都置為〇if(a[30]%2==0)bei[2]=l!〃判斷能否被2整除sum=0;for(i=0;i<N;i++)sum+=a[i]: 〃計算各位數(shù)字之和if(sum%3==0)bei[3]=l: 〃判斷能否被3整除if(sum%9==0)bei[9]=l; 〃判斷能否被9整除if((a[29]*10+a[30])%4==0)bei[4]=l! 〃判斷能否被4整除if((a[28]*100+a[29]*10+a[30])%8==0)bei[8]=lJ〃判斷能否被8整除if(a[30]==0||a[30]=5)bei[5]=l; 〃判斷能否被5整除bei[6]=bei[2]&&bei[3]; 〃判斷能否被6整除bei[7]=is7bei(a): 〃判斷能否被7整除yes=0;for(i=2;i<10;i++)if(bei[i]){printf(*%d'i);yes=l;}if(yes==O)printf(*none\n);elseprintf('\n");t=getlargenumber(a);)return0;)測試數(shù)據(jù)242346812023456880642346789-11234567897535750955522478848050562346789136694250235679151445133379-177733388878954331320002345678945678903253258888888823467887654321987654399000000002345682993887-1none22333557273964248

-1標題6.8不吉利日期選做題?描述在國外,每月的13號和每周的星期5都是不吉利的。特別是當13號那天恰好是星期5時,更不吉利。已知某年的一月一日是星期w,并且這ー年一定不是閏年,求出這一年所有13號那天是星期5的月份,按從小到大的順序輸出月份數(shù)字。（w=1..7）輸入數(shù)據(jù)輸入有一行,即一月一日星期幾（w）。（1<=w<=7）輸出數(shù)據(jù)輸出有一到多行，每行ー個月份，表示該月的13日是星期五。輸入示例7輸出示例110#include<stdio.h>intmain()(intday[13]={0,12,31,28,31,30,31,30,31,31,30,31,30);intw;inti;scanf(*%d*,&w);for(i=l;i<=12;i++){w=(w+day[i])%7;if(w=5)printf(*%d\n*,i);)return0：)測試數(shù)據(jù)7110586536291242311147標題6.9計算nI選做題?描述n大于12,小于300輸入數(shù)據(jù)輸入ー個大于12的整數(shù)

輸出數(shù)據(jù)輸出有一到多行,每行輸出n!的40位輸入示例80輸出示例71569457046263802294811533723186532165584657342365752577109445058227039255480148842668944867280814080000000000000000000intmain(){inta[2800]={l,1};intn,i,j,jinwei,chenji,weishu=l;scanf(*%d*,&n);for(i=2;i<=n;i++){jinwei=O;for(j=l;j<=weishu+l||jinwei;j++){chenji=a[j]*i+jinwei;a[j]=chenji%10;jinwei=chenji/10;)while(a[j]==O)j-;weishu=j;)for(i=weishu,j=0;i>0;i—){printfa[i]);j++;if(j%40==0)printfCr\n*);)printf("\n");return0;)測試數(shù)據(jù)503041409320171337804361260816606476884437764156896051200000000000018020089606249913429965695133689846683891754034079886777794043533516004486095339598094118013811209730973563159410103739960967103213218633149527360959853196673097294565355881980647506435385685815744504080920956035846331964466489111425643001782414179675381819233864230269332781873198603960320000000000000000000000000000000000000000000029834120123992914229212415881972939646531615251301402118106664073719005187645118937287125450548771189894084631871650136718936930201533891098576013704337455357841831156848413818796433446177901477388723997

32055260908659852009635695610527306189571643799669887511192235112881115420579495836357534252370562420514255057696413664574065573507803305022276103124054810481440095278364995676990353731594942399761243108920291747984045154445397297156190514229734193476368810311459895614048441864360723929567819808241411305409998000089879964463627468013568000000000000000000000000000000000000000000000000000000000000000000000000標題6.10計算鞍點 |選做題?描述找出ー個二維數(shù)組中的鞍點,即該位置上的元素在該行上最大,在該列上最小。也可能沒有鞍點。輸入數(shù)據(jù)輸入ー個5行6列二維數(shù)組輸出數(shù)據(jù)二個表示鞍點位置的行、列號沒有鞍點輸出none輸入示例76154322794316813711759106534711279輸出示例23#include<stdio.h>intmain(){inta[5][6];inti,j,k,hang,lie;intfind=O;for(i=0;i<5;i++)for(j=0;j<6;j++)scanf&a[i][j]);for(i=0;i<5;i++){k=0;for(j=0;j<6;j++)if(a[i][j]>a[i][k])k=j;for(j=0;j<5;j++)if(a[j][k]<a[i][k])break;if(j=5){printf(*%d%d*,i+1,k+1);find=l;})if(!find)printf("none");return0;

)測試數(shù)據(jù)56148252491153833910162018837475231276236153583761259426396772935438863308765228248113228376124594363396771935438863308765228248113none7615432-12-7-2-4-3-5681371175910653471127923標題6.11找最長的單詞 選做題?描述輸入一行字符,其中包括若干個單詞,單詞之間用空格格開輸入數(shù)據(jù)輸入一行字符輸出數(shù)據(jù)?個字符串（最長的單詞）輸入示例IHUDDDGHASDFGHGFFG8976輸出示例ASDFGHGFintmain(){inti,word=0,start=0,maxlen=0,maxstart=0;charstring[81];gets(string);for(i=0;i<strlen(string);i++)if(string[i]—')word=0;else{if(word=0) 〃ー個單詞開始{word=l;start=i;}if(i-start+l>maxlen)〃當前單詞長度是否最長{maxstart二start;maxlen=i-start+l;})for(i=maxstart;string[i]!='*&&string[i]!=‘、〇’;i++)printfstring[i]);return0;)#includeくstring.h>intmain(){inti,start="l,maxlen=0,maxend=0;charstring[81];gets(string);for(i=0;i<=strlen(string);i++)if(string[i]=''I|string[i]=‘、0') 〃遇到單詞結(jié)束{if(i-start>maxlen){maxlen=i-start;maxend=iT;}start=i+l;}for(i=maxend;string[i]!=''&&i>=0;i—);〃找單詞的第一個字母,包括下一行i=i+l;for(;string[i]!=''&&string[i]!='\0';i++)printf('%c”,string[i]);return0;}intmain(){inti,pos=0,word=0,start=0,maxlen=0;charstring[81];gets(string);for(i=0;i<=strlen(string);i++)if(string[i]=='丨|string[i]=へ(T)(if(word=l&&i-start>maxlen){pos=start;maxlen=i_start;}word=0;)elseif(word==0){word=l;start=i;}for(i=pos;string[i]!=,'&&string[i]!二’、〇';i++)printf("祝",string[i]);return0;intmain(){inti,num=0,word=0,j,k;

intlen,maxlen=O,max;charstring[81];chars[10][30];gets(string);j=-l；for(i=0;i<=strlen(string);i++)if(string[i]=**||string[i]==,\0,){k++:s[j][k]=>\0，；word=0;}elseif(word==0){word=l;num++;j++;k=0;s[j][k]=string[i];}else{k++;s[j][k]=string[i];}max=0;maxlen=strlen(s[O]);for(i=l;i<=j;i++)if(strlen(s[i])>maxlen){maxlen=strlen(s[i]);max=i;}puts(s[max]);return0;)intmain(){inti,num=0,word=0;intlen,maxlen=0,start,maxstart=0;charstring[81];gets(string);for(i=0;i<=strlen(string);i++)if(string[i]=''||string[i]='、〇'){if(word){len=i-start+l;if(len>maxlen)maxlen=len,maxstart=start;}word=0;}elseif(word==0){word=l;num++;start=i;}for(i=maxstart;string[i]&&string[i]!='';i++)printf('%c”,string[i]);return0;測試數(shù)據(jù)FFFFFFFFFHJUYT FGJJJJJJFFFFFFFFFFGHUTGFFFFHJUYTFGJJJJJJFFFFFJJJJJJFFFFF

FFFFHJUYTFGJJJJJJFFFFFJJJJJJFFFFFFGHUTGFFFFHJUYTFG FGHUTGJJJJJJFFFFFJJJJJJFFFFFDFFFFFFFGGJJOPYUIGHFFFFFFFGFFFFFFFFFHJUYT123456789FGJJJJJJFGHUTGFFFFFFFFF標題6.12字符串比較描述輸入兩個字符串,比較這兩個字符串的大小。比較時從頭開始依次比較兩個串的對應字符,如果都相同,輸出〇,否則輸出不同處兩個字符ASCII碼的差。輸入數(shù)據(jù)兩個字符串輸出數(shù)據(jù)ー個整數(shù)輸入示例asdfgasghd輸出示例-3提不3組測試數(shù)據(jù):br和bhjbr和brgbrg和br#include<stdio.h>intmain(){charsi[20],s2[20];inti,cha=0;gets(si);gets(s2);i=0;for(i=0;sl[i]&&s2[i];i++)if(sl[i]!=s2[i])break;cha=sl[i]-s2[i];printf("%d",cha);return0;)測試數(shù)據(jù)abcdefabch-4abchhfabcfhf2abcdefabcdef0def3

abcabcab99ababcgggg-99標題6.13找最長的字符串描述輸入10個字符串,比較每個字符串的長度,輸出長度最長的字符串。輸入數(shù)據(jù)10個字符串輸出數(shù)據(jù)ー個字符串輸入示例asdfasghdjhghghhyuyujhjhjhjhjjjhjhjutyuyurrruhgfghfg輸出示例jhjhjhjhjj提示如何存放10個串#include<stdio.h>intmain(){inti,pos=0;charstring[10][20];for(i=0;i<10;i++)gets(string[i]);for(i=l;i<10;i++)if(strlen(string[pos])<strlen(string[i]))pos=i;puts(string[pos]);return0;)測試數(shù)據(jù)asdfasghdjhghghhyuyuabcdefghijkjhjhjutyuyuabcdefghijk

rrruhgfghfgxasdfhjghjgjxyghdjhghghhyuyuabedjhjhjutyuyurrruhgfghfgxasdfhjghjgjxaxyghdjhghghhyuyuabedjhjhjutyuyurrruhgfghfghhhhhhhgfghfghhhhhh6.14尋找扔掉的牌描述一個人手里有2,3,4,5,6,7,8,9,10,J,Q,K,A等13張牌?，F(xiàn)在,他隨便扔掉一張牌,請根據(jù)手里還剩下的牌,輸出扔掉的是哪張牌。輸入數(shù)據(jù)一行字符串輸出數(shù)據(jù)ー個字符串輸入示例5,6,7,8,9,10,J,Q,K,A,2,3輸出示例4提示如何把輸入過程中的","去除。intmain()chars[30];inta,i,sum=O;a='1'+'2'+'3'+'4'+'5'+'6'+7'+'8'+'9'+'J'+'Q'+'K'+'A'+'0,;ggts(S)；if(strlen(s)==23)puts("10");else{for(i=0;s[i];i++)if(s[i]!=',')sum+=s[i];printf("%c",a-sum);丄return0;)解2：#include<stdio.h>#include<string.h>intmain()(chars[30];inta[15],i;charpai[15][3]={"0","1","2","3","4","5"I"6","7"I"8","9","I〇","J","Q","K","A"};char*p=s;for(i=2;i<15;i++)a[i]=O;//假設所有的牌沒有g(shù)ets(s);while(*p){if(*p>='2'&&*p<='9')a[*p-'0']=1;elseif(*p=='J')a[11]=1;elseif(*p==Q)a[12]=1;elseif(*p=='K')a[13]=1;elseif(*p=='A')a[14]=1;elseif(*p=='1'&&*(p+1)=='0')a[10]=1;P++；}for(i=2;i<15;i++)if(!a[i]){printf("%s",pai[i]);break;}return0;}3,5,7,9,A,10,6,J,Q,K,2,