这是什么逻辑呀！scanf()和printf()函数都有返回值且都是整型量。scanf()的返回值是scanf()函数正确接受到的对应型号数的个数。比如 int i,a,b,c;i=scanf("%d%d%d",&a,&b,&c);如果输入1_@_100,则i=2；如果输入12_100_-1,则i=3;如果输入@_w_A,则i=0.本题的一个完整的c程序...
（还有printf("立方和为%d",a+b+c);改为printf("立方和为%f",a+b+c);输入三个数格式错误,（你的输出格式double又写成以int格式的%d 输出错了）执行下面的就可以 include "stdio.h"include "math.h"double thre(double x);void main(){ double a,b,c;printf("请输入X,Y,Z");scanf("...
include <stdio.h> //c语言 include <math.h> int main(){ double m,x,y;printf("请输入两个数：");scanf ("%d %d",&x,&y);m=pow(double x,double y);printf("%d",m);return 0;} //我没有调试哈，有问题自己该哈 include <iostream> include <cmath> using namespace std;in...
[code=C/C++][/code]#include <string$amp;>amp;$nbsp;include <sstream$amp;>amp;$nbsp;include <iostream> include <fstream$amp;>amp;$nbsp;using namespace std;int main(){ stringstream stream;int result[10];string line;ifstream fin;fin.open("kk.txt");while(!fin.eof()){ getl...
include "stdio.h"void main(){ int x,y,z,t;printf("please input three numbers\n");scanf("%d%d%d",&x,&y,&z);if(x>y){t=x;x=y;y=t;} if(x>z){t=x;x=z;z=t;} if(y>z){t=y;y=z;z=t;} printf("Smallest is %d\nBiggest is %d\n", x,z);} ...
if(c>d)printf("最大值是:%d",c);else printf("最大值是:%d",d);} break;default:break;} } 不知道可否这样？还有一个简洁一点的 include "stdio.h"void main(){ int a,b,c,d,MAX1,MAX;scanf("%d,%d,%d,%d",&a,&b,&c,&d);MAX=(a>b?a:b);MAX1=(MAX>c?MAX:c);swi...
include <stdio.h> int countdigit(int number,int digit);int main(){ int number,digit,result;printf("Enter in:number=");scanf("%d %d",&number,&digit);result = countdigit(number,digit);printf("countdigit(%d,%d)的返回值为：%d\n",number,digit,result);return 0;} int count...
#include<stdio.h> int main(){ char ch; int i; printf("输入一个字符\n"); scanf("%c",&ch); printf("输入一个数字\n"); scanf("%d",&i); ch=ch+i; if(ch>122&&ch<(123+i)) ch=(ch-122)+96; printf("%d\n%c\n",ch,ch);} 我这个程序要实现输入一个小写... 展开 飞絮...
▉以下程序的运行结果是。main(){int a,b,c,d,x;a=c=0;b=1;d=20;if(a) d=d-10;else if(!b)if(!c) x=15;else x=25;printf(“%d\n”,d);} ▉请阅读以下程序：main(){ int s,t,a,b;scanf(“%d,%d”,&a,&b);s=1;t=1;if(a>0) s=s+1;if(a>b) t=s+t;e...
第三行 的return是不需要递归了。其实还可以简化一点的:void fun(int n, int r){ if (n>1) fun(n/2,r); //大于1时(对应的二进制数多于1位的数)递归 printf("%d",n%2); //返回前输出本次的余数}void main(){ int n,r=2; scanf("%d",&n); fun(n,r); printf("\n");} 已赞过 ...