输入两个整数,求它们相除的余数.用带参的宏来编程实验
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![输入两个整数,求它们相除的余数.用带参的宏来编程实验](/uploads/image/f/7441992-0-2.jpg?t=%E8%BE%93%E5%85%A5%E4%B8%A4%E4%B8%AA%E6%95%B4%E6%95%B0%2C%E6%B1%82%E5%AE%83%E4%BB%AC%E7%9B%B8%E9%99%A4%E7%9A%84%E4%BD%99%E6%95%B0.%E7%94%A8%E5%B8%A6%E5%8F%82%E7%9A%84%E5%AE%8F%E6%9D%A5%E7%BC%96%E7%A8%8B%E5%AE%9E%E9%AA%8C)
#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