c 输入两个整数,求他们的余数,用带参数的宏来实现用iostream

#includevoidmain(){inta,b;printf("Inputtwointegers:");scanf("%d%d",&a,&b);printf("和:a+b=%d\n",a+b);p

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=

1.main(){inta,b;scanf("%d%d",&a,&b);printf("%d%d",((int)a/b),a%b);}2.main(){inta,b;scanf("%d",&a);b=

#includeintmain(){\x09inta,b;\x09scanf("%d%d",&a,&b);\x09printf("商为%.2f余数为%d\n",1.0*a/b,a%b);\x09ret

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

#include#incluedvoidmain(){inta;intb;intc,d;a=1500,b=350;c=a/b;//商数d=a%b;//余数pritf("商数为:%d\n",c);pri

#includeintmain(){inta,b,c,d=0;scanf("%d%d",&a,&b);c=a/b;d=a%b;printf("商是:%d\n",c);printf("余数是:%d\n"

#include"stdio.h"main(){inta=0,b=0,c=0,d=0;printf("请输入两个整数：");scanf("%d,%d",&a,&b);if(b==0)printf("输

#include#includemain(){inti,x,y,n,a[10]={0};scanf("%d%d",&x,&y);for(i=x;i

print改成printf就可以了标准答案拿去吧除法进行的Int型的除法自己懂得#includeintmain(void){intnum1,num2;intx,y,z,k,j;printf("Ente

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+

voidmain(){inta,b;inty;floatx;printf("请输入2个数字：");scanf("%d%d",&a,&b);x=a/b;y=a%b;printf("商：%0.3f",x)

#include"stdio.h"intmain(intargc,_TCHAR*argv[]){inta=0;intb=0;printf("请输入数值");scanf("%d%d",&a,&b);pr

#include#defineSURPLUS(a,b)(a%b)main(){inta,b;printf("pleaseinputinteger:");scanf("%d,%d",&a,&b);pri

#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)