输入两个整数,求它们相除的余数.用带参的宏来编程实验

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

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

由于被除数=商X除数+余数.又因为被除数+商+除数+余数=99.所以（商X除数+余数）+商+除数+余数=99也就是5X除数+11+5+除数++11=996X除数+27=996X除数=99-27=72除

#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

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

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)

#includeclassdigital{private:intm_num;public:digital(intnum=0){this->m_num=num;}digital(){};intGetNu

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

商是12,余数是8,那就说明被除数是除数的12倍多8.现在知道被除数与除数的差是882,说明这882是除数的11倍多8（因为剪掉除数,少了一倍）则除数为：（822-8）/11=74被除数为：74*12

设x,yx+822=y12x+8=yx=74,y=896

设除数为x，则被除数为：12x+26，则：12x+26+x+26+12=454，          

被除数+除数=501-17-8=476又因为被除数-8=17×除数所以除数=（476-8）÷（17+1）=26所以被除数=476-26=450设除数是x,则被除数是17x+8所以17x+x+8+17+

设被除数为Y,除数为XY=7X+3Y+X+7+3=61X=6,Y=45所以除数是6

#include<iostream.h>classdigital{private:intm_num;public:digital(intnum=0){this->m_num=num;

822-8=814814÷(12-1)=7474×12+8=896答:被除数是896,除数是74.

解设除数为x,则被除数为17x+817x+8+x+17+8=50118x+33=50118x=468x=26被除数：17×26+8=450

设除数为x,则有：x+822=12x+8；11x=814；x=74；所以除数为74,被除数为896

除数=(245-8)÷(4-1)=79被除数=79+245=324