浙江省2011年秋季C語(yǔ)言二級(jí)考試上機(jī)編程試題及答案(共30頁(yè))

1、精選優(yōu)質(zhì)文檔-傾情為你奉上浙江省2011年秋季C語(yǔ)言二級(jí)考試上機(jī)編程試題及答案1三個(gè)數(shù)比較大小。#include <stdio.h>void swap(_1_) /int *pa,int *pb /*交換兩個(gè)數(shù)的位置*/ int temp; temp = *pa; *pa = *pb; *pb = temp; void main() int a,b,c,temp; scanf("%d%d%d",&a,&b,&c); if(a>b) swap(&a,&b); if(b>c) swap(&b,&c)

2、; if(_2_) /a>b swap(&a,&b);printf("%d,%d,%d",a,b,c); 2表達(dá)式求和。#include <stdio.h> #include <math.h>void main() FILE *fp; float n=1,t=1,pi=0; int i; / 從以下開始答題 i=1; while(fabs(t)>=1e-6) pi=pi+t; i= -i; n=n+2; t=i/n; fp=fopen("Design1.dat","w");fprin

3、tf(fp,"%.6f",4*pi);fclose(fp); 運(yùn)行結(jié)果：3.3字母后移循環(huán)輸出。#include <stdio.h>void main() char c; c=getchar(); if(_1_) / c>='a' && c<'v' c=c+5; else if (c>='v' && c<='z') _2_ / c=c-21; putchar(c); 4求滿足條件的數(shù)。#include <stdio.h>#inclu

4、de <math.h>void main() float y=1.05; int n=1; FILE *p; / 以下開始做答 while(!(pow(y,n)<1e6 && pow(y,n+1)>1e6) n+; p=fopen("Design2.dat","w"); fprintf(p,"%d,%.0f",n,pow(1.05,n); fclose(p); 運(yùn)行結(jié)果：283,5求滿足條件的數(shù)。#include <stdio.h>void main() int m=0,t=1,n;

5、 while( _ 1 _); / (scanf("%d",&n),n<=0) while(!(t<=n&&t*2>=n) _ 2 _ / t=t*2; m+; printf("%dn",m); 6求平面點(diǎn)間的最短距離。#include <stdio.h> #include <math.h>#define len(x1,y1,x2,y2) sqrt(x1-x2)*(x1-x2)+(y1-y2)*(y1-y2)void main() FILE *p; int i,j; float c,min

6、c; float x=1.1,3.2,-2.5,5.67,3.42,-4.5,2.54,5.6,0.97,4.65; float y=-6,4.3,4.5,3.67,2.42,2.54,5.6,-0.97,4.65,-3.33; minc=len(x0,y0,x1,y1);p=fopen("Design1.dat","w"); for(i=0;i<9;i+) for(j=i+1;j<10;j+) if(c=len(xi,yi,xj,yj)<minc) minc=c; fprintf(p,"%f",minc); fcl

7、ose(p); 運(yùn)行結(jié)果：1.7Fibonacci數(shù)列求值問(wèn)題。#include <stdio.h>_1_ / long f(int n);void main() printf("%ldn",f(30); long f(int n) if( _2_ ) / n=1 | n=2 return 1; else return f(n-1)+f(n-2); 運(yùn)行結(jié)果：8多項(xiàng)式求和問(wèn)題。#include <stdio.h> #include <math.h>void main() FILE *p; int i; float x=1.279,t=1,

8、y=0; float a10=1.1,3.2,-2.5,5.67,3.42,-4.5,2.54,5.6,0.97,4.65;p=fopen("Design2.dat","w"); y=a0 ; for(i=1;i<10;i+) t=t*x; y=y+t*ai; fprintf(p,"%f",y); fclose(p); 運(yùn)行結(jié)果：98.9整數(shù)轉(zhuǎn)換為字符串。#include<stdio.h>void itoa(long i,char *s) if(i=0) return; /* 1 */ *s = '1'

9、;+i%10;/*s='0'+i%10 itoa(i/10,s-1); void main() long n; char str7="" scanf("%ld",&n); /* 2 */ itoa(n,str+6);/itoa(n,str+5); printf("%s",str); 10Fibonacci數(shù)列求值問(wèn)題。#include<stdio.h>void main() FILE *p; int i; float f1=1.0,f2=2.0,t1=2.0,t2=3.0,s; float f,t;

10、s=t1/f1+t2/f2; p=fopen("Design1.dat","w"); for(i=3;i<40;i=i+2) t1=t1+t2; t2=t1+t2; f1=f1+f2; f2=f1+f2; s=s+t1/f1+t2/f2; fprintf(p,"%.6f",s); fclose(p); 運(yùn)行結(jié)果：65.11數(shù)組賦值。#include <stdio.h>void main() int a10,b10,i; printf("ninput 10 numbers: "); for (i=0

11、; i<10;i+) /* 數(shù)組輸入 */ scanf("%d", &ai); for (i=1; i<10; i+) bi=_1_; / bi=ai+ai-1; for (i=1; i<10; i+) printf("%3d",bi); if (_2_) printf("n"); / i%3=0 12求各點(diǎn)距離和。#include<stdio.h> #include<math.h>void main() FILE *p; int i; float x10=-1.5,2.1,6.3,3

12、.2,-0.7,7.0,5.1,3.2,4.5,7.6; float y10=3.5,7.6,8.1,4.5,6.0,1.1,1.2,2.1,3.3,4.4; float s=0.0;p=fopen("Design2.dat","w"); for(i=0;i<10;i+) s=s+sqrt(pow(xi-1, 2)+pow(yi-1, 2); fprintf(p,"%.6f",s); fclose(p); 運(yùn)行結(jié)果：52.13十進(jìn)制數(shù)轉(zhuǎn)換為二進(jìn)制數(shù)。#include <stdio.h>void dec2bin(int

13、 m) int bin32,j; for(j=0;m!=0;j+) binj= _1_; / m%2 m=m/2; for(;j!=0;j-) printf("%d", _2_ ); / binj-1 void main() int n; scanf("%d",&n); dec2bin(n); 14求符合條件的數(shù)列之和。#include <stdio.h>#include <math.h>void main() FILE *p; float s=0,a=81;int i; p=fopen("Design2.dat

14、","w"); for(i=1;i<=30;i+) s=s+a; a=sqrt(a); fprintf(p,"%.3f",s); fclose(p); 運(yùn)行結(jié)果：121.33615在字符串中刪除數(shù)字字符。#include <stdio.h> #include <string.h> #include <ctype.h>void f(char *s) int i=0; while(si!='0') if(isdigit(si) _1_(s+i,s+i+1); / strcpy _2_ i+;

15、 / else void main() char str80; gets(str); f(str); puts(str); 16求滿足條件的數(shù)。#include <stdio.h>void main() FILE *p; float f(float x,float y),min; int x,y,x1,y1; p=fopen("Design1.dat","w"); min=f(1,1); for(x=1;x<=6;x+) for(y=1;y<=6;y+) if (f(x,y)<min) x1=x;y1=y;min=f(x,y

16、); fprintf(p,"%d,%d",x1,y1); fclose(p); float f(float u,float v) return (3.14*u-v)/(u+v); 運(yùn)行結(jié)果：1, 617去除數(shù)組中的負(fù)數(shù)。#include <stdio.h>void f(int *a,int *m) int i,j; for(i=0;i<*m;i+) if(ai<0) for(j=i-;j<*m-1;j+) aj=aj+1; _1_; / *m=*m-1; void main() int i,n=7,x7=1,-2,3,4,-5,6,-7; _2

17、_; / f(x,&n); for(i=0;i<n;i+) printf("%5d",xi); printf("n"); 運(yùn)行結(jié)果：1 3 4 618二維數(shù)組中的運(yùn)算。#include <stdio.h> #include <math.h>void main() float a33=1.3,2.7,3.6,2,3,4.7,3,4,1.27; FILE *p; float x; int i,j; for(i=0;i<3;i+) x=fabs(ai0); for(j=1;j<3;j+) if(fabs(aij

18、>x) x=fabs(aij); for(j=0;j<3;j+) aij=aij/x; p=fopen("Design2.dat","w"); for(i=0;i<3;i+) for(j=0;j<3;j+) fprintf(p,"%10.6f",aij); fprintf(p,"n"); fclose(p); 運(yùn)行結(jié)果：0. 0. 1. 0. 0. 1. 0. 1. 0.19平面上各點(diǎn)距離計(jì)算。#include <stdio.h> #include <math.h>

19、#include <stdlib.h>void main() int i,n; /* 1 */ struct axy float x,y; a;/ struct axy float x; float y; *a; scanf("%d",&n); a=(float*) malloc(n*2*sizeof(float); for(i=0;i<n;i+) /* 2 */ scanf("%f%f",ai.x,ai.y); / scanf("%f%f",&ai.x,&ai.y); for(i=0;i&l

20、t;n;i+) if(sqrt(ai.x*ai.x+ai.y*ai.y)<=5) printf("%f,%fn",ai.x,ai.y); *試題本身有錯(cuò)誤，a=(struct axy *) malloc(n*2*sizeof(float);20從a數(shù)組中找出偶數(shù)放入b數(shù)組。#include <stdio.h> void main() FILE *p; int i,j,temp,n=0; int a10=7,6,20,3,14,88,53,62,10,29,b10; for(i=0;i<10;i+) if(ai%2=0) bn+=ai; for(i=0

21、;i<n-1;i+) for(j=0;j<n-i-1;j+) if(bj>bj+1) temp=bj;bj=bj+1;bj+1=temp; p=fopen("Design1.dat","w"); for(i=0;i<n;i+) fprintf(p,"%3d",bi); if ( (i+1)%3=0) fputc (p, n); fclose(p); 運(yùn)行結(jié)果：6 10 14 20 62 8821求輸入整數(shù)的各位數(shù)字之和。 #include <stdio.h> #include <math.h&

22、gt; void main() int n,s=0; scanf("%d",&n); _ 1 _ / n=fabs(n); while(n!=0) _ 2 _/ s+=n%10; n=n/10; printf("%dn",s); 22關(guān)于生產(chǎn)能力的數(shù)學(xué)應(yīng)用題。#include<stdio.h>int year(int x) float p=11.5; int y=1999; while(p<=x) p=p*(1+0.098); y+; return y; void main() FILE *p; p=fopen("de

23、sign.dat","w"); fprintf(p,"%d,%d",year(20),year(30); fclose(p); 運(yùn)行結(jié)果：2005,201023窮舉法求解方程。#include <stdio.h>void main() FILE *p; int x,y,z,k=0; p=fopen("Design1.dat","w"); for(x= -45;x<45;x+) for(y= -45;y<45;y+) for(z= -45;z<45;z+) if(x*x+y*y

24、+z*z=2000) k+; fprintf(p,"%d",k); fclose(p); 運(yùn)行結(jié)果：14424字符串排序。#include <stdio.h> #include <string.h>void main() FILE *p; char *s="634,.%w|sq2",c; int i,j,k,n=strlen(s); p=fopen("Design2.dat","w"); for(i=0;i<n-1;i+) for(j=i+1;j<n;j+) if( *(s+j)

25、<*(s+i) ) c=*(s+i); *(s+i)=*(s+j); *(s+j)=c; for(i=0;i<n;i+) fputc(si,p); fclose(p); 運(yùn)行結(jié)果：%,.2346qsw|25將整數(shù)首尾倒置。#include <stdio.h> #include <math.h>long f(long n) long m,y=0; m=fabs(n); while(m!=0) y=y*10+m%10; _1_ / m/=10; if(n>=0) return y; else _2_ / return y; void main() prin

26、tf("%ldt",f(12345); printf("%ldn",f(-34567); 運(yùn)行結(jié)果：54321 -7654326求數(shù)組的平均值，及與平均數(shù)的差。#include <stdio.h> #include <math.h>void main() FILE *p; int i,k=0;float x10=7.23,-1.5,5.24,2.1,-12.45,6.3,-5,3.2,-0.7,9.81,d,v=0;for(i=0;i<10;i+) v+=xi; v=v/10; d=fabs(x0-v);p=fopen(&q

27、uot;Design1.dat","w"); for(i=1;i<10;i+) if(fabs(xi-v)<d) d=fabs(xi-v); k=i; fprintf(p,”%.5f”,xk); fclose(p); 運(yùn)行結(jié)果：2.1000027求平方根數(shù)列之和。#include <stdio.h> #include <math.h>void main() FILE *p; int i; double s=0; for(i=2;i<=10;i+) s+=sqrt(i); p=fopen(“design2.dat”,”w”)

28、; fprintf ( p, “%.10fn”, s); fclose(p); 運(yùn)行結(jié)果：21.28求多項(xiàng)式之和#include <stdio.h>void main() int i,a,n; long t=0; /* 1 */ s=0;/ long s=0; scanf("%d%d",&a,&n); for(i=1;i<=n;i+) /* 2 */ t=t*10+i;/ t=t*10+1 s=s+t; s=s*a; printf("%ldn",s); 29計(jì)算學(xué)生的平均成績(jī)，并輸出。#include <stdio

29、.h>struct STUDENT char name16; int math; int english; int computer; int average; ;void GetAverage(struct STUDENT *pst) /* 計(jì)算平均成績(jī) */ int sum=0; sum = _1_ /sum+pst->math+pst->english+pst->computer; pst->average = sum/3; void main() int i; struct STUDENT st4="Jessica",98,95,90,

30、"Mike",80,80,90, "Linda",87,76,70,"Peter",90,100,99; for(i=0;i<4;i+) GetAverage (_2_); / st+i printf("NametMathtEnglishtComputAveragen"); for(i=0;i<4;i+) printf("%st%dt%dt%dt%dn",,sti.math,sti.english, puter,sti.average); 30求符合條件

31、的數(shù)。#include <stdio.h> #include <math.h> #include <stdlib.h>void main( ) FILE *p; int i,j; (p=fopen("design.dat","w"); for(i=1; ; i+) if(i%3=1&&i%5=3&&i%7=5&&i%9=7) break; fprintf(p,"%d",i); fclose(p); 運(yùn)行結(jié)果：31331求Armstrong數(shù)。#incl

32、ude<stdio.h> #include<math.h>void main() int i, m,s=0;printf("armstrong numbers in 100-999:");for(i=100; i<1000; i+)m=i; s=0; while (m!=0) s+=pow(m%10,3); m=m/10; if(s= =i) printf("%5d", i); 運(yùn)行結(jié)果：153 370 371 40732將兩個(gè)字符串連接起來(lái)。#include <stdio.h>void main() char

33、s180,s240; int i=0,j=0; printf("ninput the first string:"); scanf("%s",s1); printf("ninput the second string:"); scanf("%s",s2); while (s1i !='0') /* 1 */ i+1;/ i+; while (s2j !='0') /* 2 */ s1+i=s2+j; / s1i+=s2j+; /* 拼接字符到s1 */ s1i ='0'

34、; printf("nnew string: %s",s1); 33選擇法排序。#include <stdio.h> #define N 10void main() int i,j,min,temp; int aN=5,4,3,2,1,9,8,7,6,0; printf("nThe array is:n"); /* 輸出數(shù)組元素 */ for (i=0;i<N;i+) printf("%5d",ai); for (i=0;i<N-1;i+) /* 排序操作 */ min = i; for (j=i+1; j&l

35、t;N; j+) /* 1 */ if (amin<=aj) min =j; / if (amin>=aj) min =j; /* 2 */ temp=amin; amin=aj; aj=temp; /* 數(shù)據(jù)交換 */ temp=amin; amin=ai; ai=temp; printf("nThe sorted numbers: n"); /* 輸出排序結(jié)果 */ for (i=0;i<N;i+) printf("%5d",ai); printf("n"); 34計(jì)算字符串中字符權(quán)重值。#include<

36、stdio.h> #include<math.h>void main() FILE *p; int i,w; char *s="we45*&y3r#$1" p=fopen(“Design1.dat”,”w”); for(i=0;si!=0;i+) w=si*(i+1); fprintf(p,“%d”,w); fclose(p); 運(yùn)算結(jié)果：658835將字符串中的某個(gè)字符刪除。#include <stdio.h>void main() char s80; int i,j; gets(s); for(i=j=0;_1_;i+) / si!

37、=0 if(si != 'c') sj=si; _2_ /j+; sj='0' puts(s); 36計(jì)算表達(dá)式值。#include <stdio.h>void main() FILE *p; long s=1,k=1; int i;for(i=2;i<=12;i+) k*=i; s+=k; p=fopen(“Design2.dat”,”w”); fprintf(p,”%ld”,s); fclose(p); 運(yùn)算結(jié)果：37求滿足表達(dá)式要求的最小值。#include<stdio.h> #include<math.h>voi

38、d main() FILE *p; int x,y,x1,y1; float z,z1;p=fopen("Design1.dat","w");z1=10*cos(0-4)+5*sin(0-2); for(x=0;x<=10;x+)for(y=0;y<=10;y+)z=10*cos(x-4)+5*sin(y-2);if(z<z1)z1=z;x1=x;y1=y; fprintf(p,”%d,%d”,x1,y1);fclose(p); 運(yùn)算結(jié)果：1, 738計(jì)算親密數(shù)對(duì)。#include <stdio.h>void main()

39、FILE *p; int a,b,c,k; p=fopen("Design1.dat","w"); for(a=6;a<=5000;a+) for(k=1,b=0;k<=a/2;k+) if(a%k= =0) b+=k; for(k=1,c=0;k<=b/2;k+) if(b%k= =0) c+=k; if(a= =c&&a!=b) fprintf(p,"%6d,%6dn",a,b); fclose(p); 運(yùn)行結(jié)果：220, 284 284, 220 1184, 1210 1210, 1184 26

40、20, 2924 2924, 262039十進(jìn)制轉(zhuǎn)換為十六進(jìn)制數(shù)。# include <stdio.h> # include <string.h>char trans(int x)if(x<10) return '0'+x;/*1*/else return 'a'+x; / else return 'a'+(x-10); int DtoH(int n,char *str)int i=0;while(n!=0)stri=trans(n%16); /*2*/n%=16; / n/=16;i+;return i-1; vo

41、id main()int i,k,n; char *str;scanf("%d",&n); k=DtoH(n,str);for (i=0;i<=k;i+) printf("%c",strk-i); 40將字符串中的所有非英文字母刪除后輸出。#include <stdio.h> #include <string.h>void main() char str256; int i,j,k=0,n; gets(str); n=strlen(str); for(i=0;i<n;i+) /*1*/if (tolower(s

42、tri)<'a' | tolower(stri)>'z') /if (tolower(stri)>='a' &&tolower(stri)<'=z') /*2*/ strn=stri; n+; / strk=stri; k+; strk='0' printf("%sn",str); 題目有錯(cuò)，程序開頭必須加上ctype.h 41輸出整數(shù)的質(zhì)數(shù)因子。#include <stdio.h>void main() int n,i; scanf(&quo

43、t;%d",&n); /* 1 */ i=1; / i=2; while(n>1) if(n%i= =0) printf("%dt",i); n/=i; else /* 2 */ n+; / i+; 42計(jì)算整數(shù)各位數(shù)字之和。#include <stdio.h> #include <math.h>void main() int n,s=0; scanf("%d",&n); n=fabs(n); /* 1 */ while(n>1) / while(n!=0) s=s+n%10; /* 2 */

44、 n=n%10; / n=n/10; printf("%dn",s); 43在字符串中查找單詞。#include <stdio.h>int find(char *str) char *fstr="the" int i=0,j,n=0; while (stri!='0')for(_1_) / j=0;fstrj!=0;j+if (strj+i!=fstrj) break;if (_2_) n+; / fstrj=0i+; return n; void main() char a80; gets(a); printf("

45、%d",find(a); 44逆序連接字符串。#include <stdio.h> #include <string.h>void fun(char *s,char *t) int i,sl; sl=strlen(s); for(i=0;i<sl;i+) ti=si; for(i=0;i<sl;i+) /*1*/ tsl+i=ssl-i; / t2*sl - 1=si; /*2*/ tsl="0" / t2*sl=0; void main() char s100,t100; scanf("%s",s); fu

46、n(s,t); printf("%s",t); 45統(tǒng)計(jì)字符串中英文字母、數(shù)字、其他字符出現(xiàn)的次數(shù)。#include <stdio.h> #include <ctype.h>void main() char a80; int n3=0,i; gets(a); _1_ / for(i=0;ai!=0;i+) if (tolower(ai)>='a' && tolower(ai)<='z') /*統(tǒng)計(jì)字母?jìng)€(gè)數(shù)*/ n0+; else if (_2_) /*統(tǒng)計(jì)數(shù)字個(gè)數(shù)*/ / ai>=0&

47、amp;&ai<=9n1+; else n2+; for(i=0;i<3;i+) printf("%dn",ni); 46求數(shù)列之和。#include <stdio.h> #include <math.h>void main() FILE *p; float s=1,t=1,i=3; p=fopen("design1.dat","w");while(fabs(1/t) >= 1e-5)t=-t* (i-1)*i; s=s+1/t; i+=2; fprintf(p,"%.6f&

48、quot;,s); fclose(p); 運(yùn)行結(jié)果：0. 47計(jì)算多項(xiàng)式之和。#include <stdio.h>float f(float,float*,int);void main() float b5=1.1,2.2,3.3,4.4,5.5; printf("%fn",f(1.7,b,5); float f( _1_ ) / float x, float *a, int n float y=a0,t=1; int i; for(i=1;i<n;i+) t=t*x; y=y+ai*t; _2_ / return y; 運(yùn)行結(jié)果：81.48求平均值，并將

49、大于平均值的元素進(jìn)行求和。 #include <stdio.h> void main() FILE *p; int i; float a10=7.23,1.5,5.24,2.1,2.45,6.3,5,3.2,0.7,9.81,sum,v=0;p=fopen("Design2.dat","w");for(i=0;i<10;i+)v=v+ai; v/=10;for(sum=0,i=0;i<10;i+)if(ai>=v) sum=sum+ai;fprintf(p ," %.5f", sum); fclose(p

50、); 運(yùn)行結(jié)果：33.5800049統(tǒng)計(jì)在圓內(nèi)的點(diǎn)的個(gè)數(shù)。#include <stdio.h> #include <math.h>void main() FILE *p; int i,k=0; float x=1.1,3.2,-2.5,5.67,3.42,-4.5,2.54,5.6,0.97,4.65; float y=-6,4.3,4.5,3.67,2.42,2.54,5.6,-0.97,4.65,-3.33; for(i=0;i<10;i+) if(pow(xi-1,2)+pow(yi+0.5,2)<=25) k+; p=fopen("desi

51、gn2.dat","w"); fprintf(p,"%d",k); fclose(p); 運(yùn)行結(jié)果：350計(jì)算素?cái)?shù)個(gè)數(shù)，及素?cái)?shù)之和。#include<stdio.h>int cnt,sum;void JSValue() int i,j,half; for(i=500;i<=800;i+) half=i/2; for(j=2;j<=half;j+) if(i%j=0) break; if(j>half) cnt+;sum+=i; void main() FILE *p; p=fopen("design.da

52、t","w"); JSValue(); fprintf(p,"%d,%d",cnt,sum); fclose(p); 運(yùn)行結(jié)果：44,2854251求Fibonacci數(shù)列問(wèn)題。方法二： int f1, f2, i; f1=f2=1; for(i=2;i+) f1=f1+f2; f2=f1+f2; if(f1>t | f2>t) break; if (f1>t && f2>t)return f1;if (f1<t && f2>t) return f2;#include<stdio.h>int JSValue(int t) int F40;int i; F0=0;F1=1; for(i=2;i+) Fi=Fi-1+Fi-2; if(Fi>t) break; return Fi; void main() FILE *p; p=fopen("design.dat","w"); fprintf