C語言基本算法

.經(jīng)典C語言算法2009年08月17日星期日16:51C語言的學(xué)習(xí)要從基礎(chǔ)開始，這里是100個經(jīng)典的算法－１C語言的學(xué)習(xí)要從基礎(chǔ)開始，這里是100個經(jīng)典的算法題目：古典問題：有一對兔子，從出生后第3個月起每個月都生一對兔子，小兔子長到第三個月后每個月又生一對兔子，假如兔子都不死，問每個月的兔子總數(shù)為多少？__________________________________________________________________程序分析：兔子的規(guī)律為數(shù)列1,1,2,3,5,8,13,21....___________________________________________________________________程序源代碼：main(){longf1,f2;inti;f1=f2=1;for(i=1;i<=20;i++){printf("%12ld%12ld",f1,f2);if(i%2==0)printf("\n");/*控制輸出，每行四個*/f1=f1+f2;/*前兩個月加起來賦值給第三個月*/f2=f1+f2;/*前兩個月加起來賦值給第三個月*/}}上題還可用一維數(shù)組辦理，youtry!題目：判斷101-200之間有多少個素數(shù)，并輸出所有素數(shù)。__________________________________________________________________..程序分析：判斷素數(shù)的方法：用一個數(shù)分別去除2到sqrt(這個數(shù))，假如能被整除，則表示此數(shù)不是素數(shù)，反之是素數(shù)。___________________________________________________________________程序源代碼：#include"math.h"main(){intm,i,k,h=0,leap=1;printf("\n");for(m=101;m<=200;m++){k=sqrt(m+1);for(i=2;i<=k;i++)if(m%i==0){leap=0;break;}if(leap){printf("%-4d",m);h++;if(h%10==0)printf("\n");}leap=1;}printf("\nThetotalis%d",h);}題目：打印出所有的“水仙花數(shù)”，所謂“水仙花數(shù)”是指一個三位數(shù)，其各位數(shù)字立方和等于該數(shù)自己。比方：153是一個“水仙花數(shù)”，因為153=1的三次方＋5的三次方＋3的三次方。__________________________________________________________________程序分析：利用for循環(huán)控制100-999個數(shù)，每個數(shù)分解出個位，十位，百位。___________________________________________________________________..程序源代碼：main(){inti,j,k,n;printf("'waterflower'numberis:");for(n=100;n<1000;n++){i=n/100;/*分解出百位*/j=n/10%10;/*分解出十位*/k=n%10;/*分解出個位*/if(i*100+j*10+k==i*i*i+j*j*j+k*k*k){printf("%-5d",n);}}printf("\n");}題目：將一個正整數(shù)分解質(zhì)因數(shù)。比方：輸入90,打印出90=2*3*3*5。__________________________________________________________________程序分析：對n進行分解質(zhì)因數(shù)，應(yīng)先找到一個最小的質(zhì)數(shù)k，此后按下述步驟完成：假如這個質(zhì)數(shù)恰等于n，則說明分解質(zhì)因數(shù)的過程已經(jīng)結(jié)束，打印出即可。假如n<>k，但n能被k整除，則應(yīng)打印出k的值，并用n除以k的商,作為新的正整數(shù)你n,重復(fù)履行第一步。假如n不可以被k整除，則用k+1作為k的值,重復(fù)履行第一步。___________________________________________________________________程序源代碼：/*zhengintisdividedyinshu*/main(){intn,i;printf("\npleaseinputanumber:\n");scanf("%d",&n);printf("%d=",n);for(i=2;i<=n;i++){..while(n!=i){if(n%i==0){printf("%d*",i);n=n/i;}elsebreak;}}printf("%d",n);}題目：利用條件運算符的嵌套來達成本題：學(xué)習(xí)成績>=90分的同學(xué)用A表示，60-89分之間的用B表示，60分以下的用C表示。__________________________________________________________________程序分析：(a>b)?a:b這是條件運算符的基本例子。___________________________________________________________________程序源代碼：main(){intscore;chargrade;printf("pleaseinputascore\n");scanf("%d",&score);grade=score>=90?'A'score>=60?'B':'C');printf("%dbelongsto%c",score,grade);}題目：輸入兩個正整數(shù)m和n，求其最大合約數(shù)和最小公倍數(shù)。__________________________________________________________________..程序分析：利用輾除法。___________________________________________________________________程序源代碼：main(){inta,b,num1,num2,temp;printf("pleaseinputtwonumbers:\n");scanf("%d,%d",&num1,&num2);if(num1{temp=num1;num1=num2;num2=temp;}a=num1;b=num2;while(b!=0)/*利用輾除法，直到b為0為止*/{temp=a%b;a=b;b=temp;}printf("gongyueshu:%d\n",a);printf("gongbeishu:%d\n",num1*num2/a);}題目：輸入一行字符，分別統(tǒng)計出其中英文字母、空格、數(shù)字和其余字符的個數(shù)。__________________________________________________________________程序分析：利用while語句,條件為輸入的字符不為'\n'.___________________________________________________________________程序源代碼：#include"stdio.h"main()..{charc;intletters=0,space=0,digit=0,others=0;printf("pleaseinputsomecharacters\n");while((c=getchar())!='\n'){if(c>='a'&&c<='z'||c>='A'&&c<='Z')letters++;elseif(c=='')space++;elseif(c>='0'&&c<='9')digit++;elseothers++;}printf("allinall:char=%dspace=%ddigit=%dothers=%d\n",letters,space,digit,others);}題目：求s=a+aa+aaa+aaaa+aa...a的值，其中a是一個數(shù)字。比方2+22+222+2222+22222(此時共有5個數(shù)相加)，幾個數(shù)相加有鍵盤控制。__________________________________________________________________程序分析：要點是計算出每一項的值。___________________________________________________________________程序源代碼：main(){inta,n,count=1;longintsn=0,tn=0;printf("pleaseinputaandn\n");scanf("%d,%d",&a,&n);printf("a=%d,n=%d\n",a,n);while(count<=n){tn=tn+a;..sn=sn+tn;a=a*10;++count;}printf("a+aa+...=%ld\n",sn);}題目：一個數(shù)假如恰巧等于它的因子之和，這個數(shù)就稱為“完數(shù)”。比方6=1＋2＋3.編程找出1000之內(nèi)的所有完數(shù)。___________________________________________________________________程序源代碼：main(){staticintk［10］;inti,j,n,s;for(j=2;j<1000;j++){n=-1;s=j;for(i=1;i{if((j%i)==0){n++;s=s-i;k［n］=i;}}if(s==0){printf("%disawanshu",j);for(i=0;iprintf("%d,",k);printf("%d\n",k［n］);}}}題目：一球從100米高度自由落下，每次落地后反跳回原高度的一半；再落下，求它在第10次落地時，共經(jīng)過多少米？第10次反彈多高？..___________________________________________________________________程序源代碼：main(){floatsn=100.0,hn=sn/2;intn;for(n=2;n<=10;n++){sn=sn+2*hn;/*第n次落地時共經(jīng)過的米數(shù)*/hn=hn/2;/*第n次反跳高度*/}printf("thetotalofroadis%f\n",sn);printf("thetenthis%fmeter\n",hn);}題目：一只猴子摘了N個桃子第一天吃了一半又多吃了一個,次日又吃了余下的一半又多吃了一個,到第十天的時候發(fā)現(xiàn)還有一個.___________________________________________________________________程序源代碼：/*猴子吃桃問題*/main(){inti,s,n=1;for(i=1;i<10;i++){s=(n+1)*2n=s;}printf("第一天共摘了%d個桃\n",s);}迭代法求方程根___________________________________________________________________/*迭代法求一個數(shù)的平方根*/..#defineEpsilon1.0E-6/*控制解的精度*/#include<math.h>main(){floata,x0,x1;printf("請輸入要求的數(shù):");scanf("%f",&a);x0=a/2;x1=(x0+a/x0)/2;while(fabs(x1-x0)>=Epsilon){x0=x1;x1=(x0+a/x0)/2;}printf("%f的平方根：%f.5\n",x1);}/*上題的另一種算法*/#defineEpsilon1.0E-6/*控制解的精度*/#include<stdio.h>#include<math.h>main(){floatnum,pre,this;do{scanf("%f",&num);/*輸入要求平方根的數(shù)*/}while(num<0);if(num==0)printf("therootis0");else{this=1;do{pre=this;this=(pre+num/pre)/2;}while(fabs(pre-this)>Epsilon);/*用解的精度，控制循環(huán)次數(shù)*/}printf("therootis%f",this);}用牛頓迭代法求方程2*x*x*x-4*x*x+3*x-6的根../*牛頓迭代法*/#defineEpsilon1.0E-6/*控制解的精度*/#include<math.h>main(){floatx1,x0=1.5;x1=x0-(2*x0*x0*x0-4*x0*x0+3*x0-6)/(6*x0*x0-8*x0+3);while(fabs(x1-x0>=Epsilon){x0=x1;x1=x0-(2*x0*x0*x0-4*x0*x0+3*x0-6)/(6*x0*x0-8*x0+3);}printf("方程的根為%f\n",x1);}用二分法求上題/*二分法*/#defineEpsilon1.0E-5/*控制解的精度*/#include<math.h>main(){folatx1,x2,x0,f1,f2,f0;x0=(x1+x2)/2;f0=2*x0*x0*x0-4*x0*x0+3*x0-6;/*求中點的函數(shù)值*/while(fabs(f0)>=Epsilon){if(f0*f1<0){x2=x0;f2=2*x2*x2*x2-4*x2*x2+3*x2-6;}if(f0*f2<0){x1=x0;f1=2*x1*x1*x1-4*x1*x1+3*x1-6;}x0=(x1+x2)/2;f0=2*x0*x0*x0-4*x0*x0+3*x0-6;}printf("用二分法求得方程的根：%f\n",x0);}..題目：打印出以以下圖案（菱形）****___________________________________________________________________程序分析：先把圖形分紅兩部分來對待，前四行一個規(guī)律，后三行一個規(guī)律，利用雙重for循環(huán)，第一層控制行，第二層控制列。___________________________________________________________________程序源代碼：main(){inti,j,k;for(i=0;i<=3;i++){for(j=0;j<=2-i;j++)printf("");for(k=0;k<=2*i;k++)printf("*");printf("\n");}for(i=0;i<=2;i++){for(j=0;j<=i;j++)printf("");for(k=0;k<=4-2*i;k++)printf("*");printf("\n");}}題目：一個5位數(shù)，判斷它能否是回文數(shù)。即12321是回文數(shù)，個位與萬位相同，..十位與千位相同。___________________________________________________________________程序分析：同29例___________________________________________________________________程序源代碼：main(){longge,shi,qian,wan,x;scanf("%ld",&x);wan=x/10000;qian=x%10000/1000;shi=x%100/10;ge=x%10;if(ge==wan&&shi==qian)/*個位等于萬位并且十位等于千位*/printf("thisnumberisahuiwen\n");elseprintf("thisnumberisnotahuiwen\n");}題目：請輸入星期幾的第一個字母來判斷一下是星期幾，假如第一個字母相同，則連續(xù)判斷第二個字母。___________________________________________________________________程序分析：用狀況語句比較好，假如第一個字母相同，則判斷用狀況語句或if語句判斷第二個字母。___________________________________________________________________程序源代碼：#include<stdio.h>voidmain(){charletter;printf("pleaseinputthefirstletterofsomeday\n");..while((letter=getch())!='Y')/*當(dāng)所按字母為Y時才結(jié)束*/{switch(letter){case'S':printf("pleaseinputsecondletter\n");if((letter=getch())=='a')printf("saturday\n");elseif((letter=getch())=='u')printf("sunday\n");elseprintf("dataerror\n");break;case'F':printf("friday\n");break;case'M':printf("monday\n");break;case'T':printf("pleaseinputsecondletter\n");if((letter=getch())=='u')printf("tuesday\n");elseif((letter=getch())=='h')printf("thursday\n");elseprintf("dataerror\n");break;case'W':printf("wednesday\n");break;default:printf("dataerror\n");}}}題目：Pressanykeytochangecolor,doyouwanttotryit.Pleasehurryup!___________________________________________________________________程序源代碼：#include<conio.h>voidmain(void){intcolor;for(color=0;color<8;color++){textbackground(color);/*設(shè)置文本的背景顏色*/cprintf("Thisiscolor%d\r\n",color);cprintf("ressanykeytocontinue\r\n");getch();/*輸入字符看不見*/}..}題目：學(xué)習(xí)gotoxy()與clrscr()函數(shù)___________________________________________________________________程序源代碼：#include<conio.h>voidmain(void){clrscr();/*清屏函數(shù)*/textbackground(2);gotoxy(1,5);/*定位函數(shù)*/cprintf("Outputatrow5column1\n");textbackground(3);gotoxy(20,10);cprintf("Outputatrow10column20\n");}題目：練習(xí)函數(shù)調(diào)用___________________________________________________________________程序源代碼：#include<stdio.h>voidhello_world(void){printf("Hello,world!\n");}voidthree_hellos(void){intcounter;for(counter=1;counter<=3;counter++)hello_world();/*調(diào)用此函數(shù)*/}voidmain(void){three_hellos();/*調(diào)用此函數(shù)*/}題目：文本顏色設(shè)置..___________________________________________________________________程序源代碼：#include<conio.h>voidmain(void){intcolor;for(color=1;color<16;color++){textcolor(color);/*設(shè)置文本顏色*/cprintf("Thisiscolor%d\r\n",color);}textcolor(128+15);cprintf("Thisisblinking\r\n");}題目：求100之內(nèi)的素數(shù)___________________________________________________________________程序源代碼：#include<stdio.h>#include"math.h"#defineN101main(){inti,j,line,a［N］;for(i=2;i<N;i++)a=i;for(i=2;i<sqrt(N);i++)for(j=i+1;j<N;j++){if(a!=0&&a［j］!=0)if(a［j］%a==0)［j］=0;}printf("\n");for(i=2,line=0;i<N;i++){if(a!=0){printf("%5d",a);line++;}if(line==10)..{printf("\n");line=0;}}}題目：對10個數(shù)進行排序___________________________________________________________________程序分析：可以利用選擇法，即從后9個比較過程中，選擇一個最小的與第一個元素互換，下次類推，即用第二個元素與后8個進行比較，并進行互換。程序源代碼：#defineN10main(){inti,j,min,tem,a［N］;/*inputdata*/printf("pleaseinputtennum:\n");for(i=0;i<N;i++){printf("a［%d］=",i);scanf("%d",&a);}printf("\n");for(i=0;i<N;i++)printf("%5d",a);printf("\n");/*sorttennum*/for(i=0;i<N-1;i++){min=i;for(j=i+1;j<N;j++)if(a［min］>a［j］)min=j;tem=a;a=a［min］;a［min］=tem;}/*outputdata*/printf("Aftersorted\n");for(i=0;i<N;i++)printf("%5d",a);}..題目：求一個3*3矩陣對角線元素之和___________________________________________________________________程序分析：利用雙重for循環(huán)控制輸入二維數(shù)組，再將a累加后輸出。___________________________________________________________________程序源代碼：main(){floata［3］［3］,sum=0;inti,j;printf("pleaseinputrectangleelement:\n");for(i=0;i<3;i++)for(j=0;j<3;j++)scanf("%f",&a［j］);for(i=0;i<3;i++)sum=sum+a;printf("duijiaoxianheis%6.2f",sum);}題目：有一個已經(jīng)排好序的數(shù)組?，F(xiàn)輸入一個數(shù)，要求按原來的規(guī)律將它插入數(shù)組中。___________________________________________________________________程序分析：第一判斷此數(shù)能否大于最后一個數(shù)，此后再考慮插入中間的數(shù)的狀況，插入后此元素今后的數(shù)，依次后移一個地點。___________________________________________________________________程序源代碼：main(){inta［11］={1,4,6,9,13,16,19,28,40,100};inttemp1,temp2,number,end,i,j;..printf("originalarrayis:\n");for(i=0;i<10;i++)printf("%5d",a);printf("\n");printf("insertanewnumber:");scanf("%d",&number);end=a［9］;if(number>end)a［10］=number;else{for(i=0;i<10;i++){if(a>number){temp1=a;a=number;for(j=i+1;j<11;j++){temp2=a［j］;［j］=temp1;temp1=temp2;}break;}}}for(i=0;i<11;i++)printf("%6d",a);}題目：將一個數(shù)組逆序輸出。___________________________________________________________________程序分析：用第一個與最后一個互換。___________________________________________________________________程序源代碼：#defineN5main(){inta［N］={9,6,5,4,1},i,temp;printf("\noriginalarray:\n");for(i=0;i<N;i++)..printf("%4d",a);for(i=0;i<N/2;i++){temp=a;a=a［N-i-1］;a［N-i-1］=temp;}printf("\nsortedarray:\n");for(i=0;i<N;i++)printf("%4d",a);}題目：學(xué)習(xí)static定義靜態(tài)變量的用法___________________________________________________________________程序源代碼：#include"stdio.h"varfunc(){intvar=0;staticintstatic_var=0;printf("\40:varequal%d\n",var);printf("\40:staticvarequal%d\n",static_var);printf("\n");var++;static_var++;}voidmain(){inti;for(i=0;i<3;i++)varfunc();}題目：學(xué)習(xí)使用auto定義變量的用法___________________________________________________________________程序源代碼：#include"stdio.h"main(){inti,num;num=2;for(i=0;i<3;i++)..{printf("\40:Thenumequal%d\n",num);num++;{autointnum=1;printf("\40:Theinternalblocknumequal%d\n",num);num++;}}}C語言的學(xué)基礎(chǔ)，100個經(jīng)典的算法－２程序源代碼：#include"stdio.h"main(){inti,num;num=2;for(i=0;i<3;i++){printf("\40:Thenumequal%d\n",num);num++;{staticintnum=1;printf("\40:Theinternalblocknumequal%d\n",num);num++;}}}題目：學(xué)習(xí)使用external的用法。___________________________________________________________________程序源代碼：#include"stdio.h"inta,b,c;voidadd(){inta;a=3;c=a+b;}voidmain()..{a=b=4;add();printf("Thevalueofcisequalto%d\n",c);}題目：學(xué)習(xí)使用register定義變量的方法。___________________________________________________________________程序源代碼：voidmain(){registerinti;inttmp=0;for(i=1;i<=100;i++)tmp+=i;printf("Thesumis%d\n",tmp);}題目：宏#define命令練習(xí)(1)___________________________________________________________________程序源代碼：#include"stdio.h"#defineTRUE1#defineFALSE0#defineSQ(x)(x)*(x)voidmain(){intnum;intagain=1;printf("\40:Programwillstopifinputvaluelessthan50.\n");while(again){printf("\40leaseinputnumber==>");scanf("%d",&num);printf("\40:Thesquareforthisnumberis%d\n",SQ(num));if(num>=50)again=TRUE;elseagain=FALSE;}..}題目：宏#define命令練習(xí)(2)___________________________________________________________________程序源代碼：#include"stdio.h"#defineexchange(a,b){\/*宏定義中贊同包含兩道衣裳命令的狀況，此時必定在最右邊加上"\"*/intt;\t=a;\a=b;\b=t;\}voidmain(void){intx=10;inty=20;printf("x=%d;y=%d\n",x,y);exchange(x,y);printf("x=%d;y=%d\n",x,y);}題目：宏#define命令練習(xí)(3)___________________________________________________________________程序源代碼：#defineLAG>#defineSMA<#defineEQ==#include"stdio.h"voidmain(){inti=10;intj=20;if(iLAGj)printf("\40:%dlargerthan%d\n",i,j);elseif(iEQj)printf("\40:%dequalto%d\n",i,j);elseif(iSMAj)printf("\40:%dsmallerthan%d\n",i,j);else..printf("\40:Nosuchvalue.\n");}題目：#if#ifdef和#ifndef的綜合應(yīng)用。___________________________________________________________________程序源代碼：#include"stdio.h"#defineMAX#defineMAXIMUM(x,y)(x>y)?x:y#defineMINIMUM(x,y)(x>y)?y:xvoidmain(){inta=10,b=20;#ifdefMAXprintf("\40:Thelargeroneis%d\n",MAXIMUM(a,b));#elseprintf("\40:Theloweroneis%d\n",MINIMUM(a,b));#endif#ifndefMINprintf("\40:Theloweroneis%d\n",MINIMUM(a,b));#elseprintf("\40:Thelargeroneis%d\n",MAXIMUM(a,b));#endif#undefMAX#ifdefMAXprintf("\40:Thelargeroneis%d\n",MAXIMUM(a,b));#elseprintf("\40:Theloweroneis%d\n",MINIMUM(a,b));#endif#defineMIN#ifndefMINprintf("\40:Theloweroneis%d\n",MINIMUM(a,b));#elseprintf("\40:Thelargeroneis%d\n",MAXIMUM(a,b));#endif}題目：#include的應(yīng)用練習(xí)___________________________________________________________________程序源代碼：..test.h文件以下：#defineLAG>#defineSMA<#defineEQ==#include"test.h"/*一個新文件50.c，包含test.h*/#include"stdio.h"voidmain(){inti=10;intj=20;if(iLAGj)printf("\40:%dlargerthan%d\n",i,j);elseif(iEQj)printf("\40:%dequalto%d\n",i,j);elseif(iSMAj)printf("\40:%dsmallerthan%d\n",i,j);elseprintf("\40:Nosuchvalue.\n");}題目：學(xué)習(xí)使用按位與&。___________________________________________________________________程序分析：0&0=0;0&1=0;1&0=0;1&1=1___________________________________________________________________程序源代碼：#include"stdio.h"main(){inta,b;a=077;b=a&3;printf("\40:Thea&b(decimal)is%d\n",b);b&=7;printf("\40:Thea&b(decimal)is%d\n",b);}題目：學(xué)習(xí)使用按位或|。..___________________________________________________________________程序分析：0|0=0;0|1=1;1|0=1;1|1=1___________________________________________________________________程序源代碼：#include"stdio.h"main(){inta,b;a=077;b=a|3;printf("\40:Thea&b(decimal)is%d\n",b);b|=7;printf("\40:Thea&b(decimal)is%d\n",b);}題目：學(xué)習(xí)使用按位異或^。___________________________________________________________________程序分析：0^0=0;0^1=1;1^0=1;1^1=0___________________________________________________________________程序源代碼：#include"stdio.h"main(){inta,b;a=077;b=a^3;printf("\40:Thea&b(decimal)is%d\n",b);b^=7;printf("\40:Thea&b(decimal)is%d\n",b);}題目：取一個整數(shù)a從右端開始的4～7位。..___________________________________________________________________程序分析：可以這樣考慮：先使a右移4位。設(shè)置一個低4位全為1,其余全為0的數(shù)?？捎脋(~0<<4)將上面二者進行&運算。___________________________________________________________________程序源代碼：main(){unsigneda,b,c,