C语言程序设计实验-答案.doc

-前言-/*非常感谢度娘以及各位网上C语言高手的支持，才能让敝人完成此文档的整理。本文档集合了本人、度娘、众网友的力量，其中代码的正确率约为90%（不正确的有标注）。为回报度娘及众网友的帮助，本文档免费下载。*/*配“电子科技大学出版社（周信东主编）”的C语言程序设计实验*/*努力吧，骚年以及学妹们！*/*整理ed by 口玉刀一 of GUET.*/=实验一 C语言程序初步=1.-已知a=5,b=4,c=6,求出s并换行#includestdio.h main() int a,b,c,s; a=5;b=4;c=6; s=(a+b+c)/3; printf(a=%d,b=%d,c=%dn,s=%d,a,b,c,s); 2.-输入一个数字求他的平方#includemain() int r,s;scanf(%d,&r);s=r*r;printf(s=%dn,s); 3.-（1）#include main() printf( *n); printf(*n); printf( *n); （2）#include main() int v; int a,b,c; /a,b,c aer sides,v is volume of cubea=3;b=3;c=5; v=a*b*c; printf(v=%dn,v);=实验二 数据类型、运算符和表达式=1.（1）-/总觉得打印结果怪怪的，DO YOU THINK SO?main() char s1=3,s2=4,s3=5; int c1=101,c2=102; printf(3%ct4%ct5%cn,s1,s2,s3); /3%c为输出3和%c printf(s1=%dts2=%dts3=%dn,s1,s2,s3);/注意哦，s1,s2,s3是char!而%d：输入输出为整形 %ld 长整型 %hd短整型 %hu无符号整形 %u %lu%s:输入输出为字符串 %c字符%f：输入输出为浮点型 %lf双精度浮点型 printf(c1=%dt%cn,c1,c1); /换码符t,表示水平制表位（horizeontal tab）,它的作用是将光标移到最接近8的倍数的位置 printf(c2=%dt%cn,c2,c2); /注意c1,c2的类型 （2）/运行结果为8.300000 %是求余数 先运算x-y，把结果转换为int型的有利于四则运算main() float x=8.3,y=4.2,s; int a=7; s=x+a%5*(int)(x-y)%2/3; printf(s=%f,s); （3）main() int i,j,p,q; i=3;j=6; p=i+;q=-j; printf(%d,%d,%d,%dn,i,j,p,q); p=i-+3;q=+j-4; printf(%d,%d,%d,%d,i,j,p,q); （4）main() int m=15,n=9,s; m*=3+2; printf(%d,m); n+=n-=n*=n; printf(%d,n); printf(%dn,(s=6*5,m+n+s); printf(%dn,s); 2.-（1）#include #define MM 40 main() int a=3,b=6,t; t=MM/(3+6); printf(%d%d,t,MM); /原来printf语句打成了print，printf语句中少了一个%d.（2）#include main() int m; double x; scanf(%d,&m); x=3.14*m*m; printf(%f,x); /注意分号，%f转义（3）main() int m=8,n=5; printf(%d,m*=m-n);printf(%d,%d,%d,m=m+1,n-1,m+n);3. -main() int a=9,b=5; a+=b; b=a-b; a-=b; printf(%d%d,a,b); 4. - /4.输入a,b,c三个变量，计算数学公式(a*b)/c main() int a,b,d; float s,c;scanf(%d%d%f,&a,&b,&c); d=a*b; s=d/c; printf(%f,s); =实验三 顺序结构和输入输出方法=1.-（1）main() int m=345,t; float n=2.56; t=2.56*100; printf(m=%d,n*100=%dn,m,t); （2）main() int a,b,c,s,v; scanf(%d%d%d,&a,&b,&c); s=a*b; v=a*b*c; printf(%d,%d,%dn,a,b,c); printf(s=%d,v=%dn,s,v); 2.-（1）main() int m,n; float x=3.5,y=6.2;scanf(%d,%d,&m,&n); printf(%6d,%6d,m,n); /*每个数的输出宽度为6,两个数之间逗号建个。*/ printf(x=%7.2f,y=%7.2f,x,y); /*个数的输出宽度7,小数位2*/ （2）main() int a,b; long m,n; float p,q; scanf(a=%d,b=%dn%f,%f,&a,&b,&p,&q); scanf(%ld,&m); scanf(%ld,&n); printf(a=%d,b=%dn m=%ld,n=%ldn,a,b,m,n); printf(p=%f,q=%fn,p,q); /打印结果有误！3.-main() int i,j,k,ss,m=1,n=1; scanf(%d%d,&i,&j); m+=i+; n*=-j; ss=(k=m+2,n-5,k+n); printf(%d,ss); 4.-#include main() char c1,c2; c1=getchar();c2=getchar(); putchar(c1);putchar(c2); printf(c1=%c,c2=%c,c1,c2); printf(%d.%dn,c1,c2); 5. -#include main() int a=790,b,c; b=790/60; c=790%60; printf(%d小时%d分钟,b,c); 6. -#include main() int a,b,c,d; scanf(%d%d%d,&a,&b,&c); d=a; a=c; c=b; b=d; printf(%d,%d,%dn,a,b,c); 7. -#include main() double a; scanf(%lf,&a); printf(%.2fn,a); =实验四 逻辑运算判断选取控制=1.-1. a. 错误原因分析：Switch分号错误 switch(a*a+b*b) case 3: case 1:y=a+b;break; case 0:y=b-a;break; b. 错误原因分析：switch语句使用错误，后跟括号。Default放在后边。 switch (a) case 10:y=a-b;break; case 11:x=a*b;break; default:x=a+b; c. 错误原因分析：没有输出语句。 switch(a+b) case 10:x=a+b;break; case 11:y=a-b;break; 2.-（1）main ( ) int x,y; scanf(%d,&x ); if (x-5&x0&x-5)&(x0&x0?a:-a;printf(abs(a)=%d,M);4.-main()int c,t,m;printf(input the number of coat and trousers your want buy:n);scanf(%d%d,&c,&t);if(c=t) if(c=50) m=c*80; else m=c*90; else if(ct)if(t=50) m=t*80+(c-t)*60; else 90*t+(c-t)*60;else if(c=50) 80*t+(t-c)*45;else m=c*90+(t-c)*45;printf(%d,m); 5. -#includeint main(void) int a, b, c;scanf(%d%d%d,&a,&b,&c); if(a + b = c | a + c = b | b + c = a) printf(不构成三角形n);else if (a = b & b = c) printf(等边三角形n); else if (a = b | b = c | a = c) printf(等腰三角形n); else printf(不等边三角形n); return 0; 6. -#includestdio.h main() long a;scanf(%ld,&a);if(a%5=0)if(a%7=0)printf(yesn|); else printf(non); =实验五 循环结构=1. -#includestdio.hmain() int n; while(1) printf(Enter a number:); scanf(%d,&n); if(n%2=1) printf(I said); continue;break;printf(Thanks.Ineeded that!);2. -#include main() int c;while ( (c=getchar () ) !=n)if ( (c=A|c=a) )c += 2;else if (c=y|c=Y|c=z|c=Z) c=c-24;printf(%c,c); putchar (n) ;3. -#includeint main() int m, k; for( m = 1; m = 4; m +) if ( m = 1 | m = 4 ) for( k = 1; k = 6; k+) printf(*);printf(n); else printf(* *n); return 0; 4.-#include main() int a,b,c; printf( * 1 2 3 4 5 6 7 8 9n);printf( -n); for(a=1;a=9;a+)printf(%3d,a); for(b=1;b=a;b+)c=a*b;printf(%3d,c);printf(n);5. -#includemain()double e=2,w=0.000001,t;double n=1,s=1;t=1/w;while(st)s=s*(n+1);e=e+1/s;n+;printf(e=%lf,n=%dn,e,(int)(n-1);6. -#include void main() int i,j,k=0,r,s=0,t=1000,a=0; for(i=2;i=t;i+) a=0; r=0; for(j=1;ji;j+) k=0; if(i%j=0) k=j; r+=k; if(r=i) a=r; if(a!=0) if(a=6) printf(%d,a); else printf(+%d,a); s+=a; printf(=%dn,s); =实验六 数组=1.-#includemain()int grade6;int i,mumber;for(i=1;i=5;+i) gradei=0;printf(enter your numbern);for(i=1;i=20;+i)scanf(%d,&mumber);if(!mumber) break;+grademumber;printf(nnresult of searchn);printf(-n);for(i=1;i=5;+i)printf(%4d %dn,i,gradei);2.-#includestdio.hmain() int n,k,xx20; int i,j,t; printf(nPlease enter a number); scanf(%d,&n); printf(nPlease enter %d numbers:,n); for(i=0;in;i+) scanf(%d,&xxi); printf(%nPlease enter another number:); scanf(%d,&k); for(i=0;ik;i+) t=xx0; for(j=1;jn;j+) xxj-1=xxj; xxn-1=t; printf(nAfter moving:n); for(i=0;in;i+) printf(%3d,xxi); printf(n);3.-#include main()int i,a100=90,180,270,380,590,530,140,750,380,b6; for(i=0;i6;i+) bi=0;for(i=0;i9;i+) switch(ai/100) case 0:b0+;break; case 1:b1+;break; case 2:b2+;break; case 3:b3+;break; case 4:b4+;break; default :b5+;break; printf(The result is:);for(i=0;i6;i+)printf(%d,bi);printf(n);4.-#include main() int i,j,t,p,q,s,n,a11=6,3,42,23,35,71,98,67,56,38; for(i=0;i10;i+) for(j=i+1;jaj) t=ai; ai=aj; aj=t; printf(%d ,ai); printf(ninput number:n); scanf(%d,&n); for(i=0;i10;i+) if(n=i;j-) aj+1=aj; break; ai=n; for(i=0;i=10;i+) printf(%d ,ai); printf(n); 5. -#includeint main() char a100, b100;int i=0, j=0, n=0;gets(a);gets(b);while(aj!=0)while(aj+i=bi)&bi!=0) i+;if(bi=0) n+;i=0;j+;printf(%d, n);=实验七 字符处理=1.-该程序的功能是：输入字符串，删除其中的数字，保留剩下的字符。#includemain()char a40,b40;int i,j; printf( Enter the string: ) ; scanf(%s,a) ; i=j=0; while (ai!=0) if (! (ai=0&ai=9) ) bj=ai; j+; +i; bj=0;printf ( %s,b) ;2.-#include string.h#include stdio.hmain()char str150=every one!; char str250=hello ; strcpy(str2+strlen(str2),str1); printf(%s,str2);3. -/本代码有些奇怪，在有的vc上可以执行，有的却不行。#include stdio.hmain()int i,pos; char str=This is a program; for(i=0;stri!=0;i+)if (stri=a) pos=i;break; printf(The position is %dn,pos);4.-#include stdio.hmain()int i,pos; char str=This is a program; for(i=0;stri!=0;i+)if (stri=a) pos=i;break; printf(The position is %dn,pos);程序运行结果如下图所示：5. -#include stdio.hmain()int i,j,k=0;char s255;while(sk=getchar()!=) k+;for(j=0;sj!=;j+)if(sj=A) sj+=32;if(sj=a) sj+=3;else if(sj=x|sj=y|sj=z)sj-=23;printf(译码后结果:);for(i=0;si!=;i+)printf(%c,si);printf(n);getchar();6. -#include stdio.hmain()int i=0;char str1255,str2255;printf(n Input string 1:);scanf(%s,str1);printf(n Input string 2:);scanf(%s,str2);while(1)if(str1i=str2i) i+;continue;elsebreak; printf(n%dn,str1i-str2i);=实验八 函数=8.1？未成功int a=3,b=5;max(int a,int b)int c;c=ab?a:b;return (c);main()extern int a,b;printf(%dn,max(a,b);8.2#include int is_prime(int m)int i;if(m=1)return 0;for(i=2;im;i+) if(m%i=0)return 0;return 1;void main() int n,c;scanf(%d,&n); c=is_prime(n); if(c=1)printf(primen);elseprintf(not primen);8.3#includeint main()int m,n,i,sum=1,t;printf(请输入m的值：n);scanf(%d,&m);printf(请输入n的值:n);scanf(%d,&n);if(m=m-n+1;i-)sum*=i;for(i=1;i=n;i+)sum/=i;printf(值为:%dn,sum);=实验九 指针=1./在每句for (j = 0; j 12; j+)前都加上p = a;这句/将if (j % 4 = 0) 改为 if (j + 1) % 4 = 0)/调试后的程序是:#includevoid main( ) int j, k, a12, *p; p = a; for (j = 0; j 12; j+) scanf(%d, p+); p = a; for (j = 0; j 12; j+) printf(%d, *p+); if (j+1) % 4 = 0) printf( n); 2. #includevoid main()void swap(int*p1,int*p2);int n1,n2,n3;int*p1,*p2,*p3;printf(input three interer n1,n2,n3:);scanf(%d%d%d,&n1,&n2,&n3);p1=&n1;p2=&n2;p3=&n3;if(n1n2)swap(p1,p2);if(n1n3)swap(p1,p3);if(n2n3)swap(p2,p3);printf(Now,the order is :%d,%d,%dn,n1,n2,n3);void swap(int *p1,int *p2)int p;p=*p1;*p1=*p2;*p2=p;3. main()int a10;int *p,i;p=a;for(i=0;i10;i+)*p=i;p+;p=a;for(i=0;i10;i+)printf(n%d=%dn,i,*p+);4.#include int table10;void lookup(int *t,int *a,int n) int k;*a=t0;for(k=1;ktk) *a=tk;void main() int k,min,*p=&min;for(k=0;k10;k+)scanf(%d,table+k);lookup(table,&min,10);printf(min=%dn,min);5.#include void main()int a55=0,*p5,i,j;for(i=0;i5;i+)pi=&