c 输入两个整数,求他们的平方和
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输入两个整数的比较:#inlucevoidmain(){inta,b;scanf("%d,%d",&a,&d);if(a>b)printf("%d",a);elseprintf("%d",b);}输入
#includevoidmach(intx,inty){printf("x²+y²=%d\n",(x*x+y*y));printf("x²-y²=%d\n",(
三个空分别填:x/2k*(a&&b)printf按位与是将两个二进制数的对应位逐一地进行逻辑与操作.与的运算规则是两个数只要有一个为0,则其逻辑与的结果就为0举个例子你就明白了!定义:a=135,b=
#includevoidmain(){floata,b,sum,change;scanf("%f,%f",&a,&b);sum=a*a+b*b;if(sum>100){change=sum/100;w
除数为0时怎样做、不为0时怎样做,不用分支结构是处理不了这种问题的,亲.再问:大哥、、、、求解。。。我郁闷了。。。再答:汗,唯一的办法就是用分支结构,if(b=0){//输出错误信息printf("除
#include"stdio.h"#include"conio.h"main(){inta,b,num1,num2,temp;printf("请输入两个整数:\n");scanf("%d%d",&nu
#includeintmain(){inta,b;floatc;printf("请输入两个整数和一个实数\n");scanf("%d%d%f",&a,&b,&c);c=a+b+c;printf("这三
#includevoidmain(){inta,b;scanf("%d%d",&a,&b);printf("%d",a*a+b*b);}
程序经过我用多组数据验证无错误,#includeintdigui(intt,intsum)//递归函数{if(t
#include#includemain(){inti,x,y,n,a[10]={0};scanf("%d%d",&x,&y);for(i=x;i
scanf("%d\n",&a);scanf("%d\n",&b);scanf是输入语句,不需要加\n然后,C语言函数最好是加一个返回值,在main前加int或者void都行,然后在函数最后加上ret
先写个函数判断n1、n2、n3大小,假设n1再问:怎么弄出n1n2n3的大小啊再答://传入三个参数n1、n2、n3的引用,程序运行完后满足n1
#includevoidmain(){inta,b,sum;scanf("%d,%d",&a,&b);sum=a+b;printf("thesumis%d",sum);}
#includeintmain(void){intcount=0,n,sum=0;whlie(scanf("%d",&n)!=EOF){if(n==-1)break;elsesum+=n;count+
#include#defineDEX(x,y)(x*x+y*y)intmain(void){inta,b;printf("输入两个数:");scanf("%d%d",&a,&b);printf("结果
设这五个连续整数为n,n+1,n+2,n+3,n+4,n2+(n+1)2+(n+2)2=(n+3)2+(n+4)2,解得n=10或n=-2,当n=10时这五个数为10,11,12,13,14,当n=-
x^2+(x+1)^2+(x+2)^2=(x+3)^2+(x+4)^2x^2-8x-20=0x=1010,11,12,13,14
usingSystem;usingSystem.Collections.Generic;usingSystem.Linq;usingSystem.Text;namespaceConsoleApplic
#include <stdio.h>int main(){ int low1,up1,low2,up2;&nbs
intnum1,num2,num3;floataverage;printf("pleasenum1,num2,num3:\n");scanf("%d,%d,%d",&num1,&num2,&num3)