C 输入两个整数a,b, 求ab的余数与ba的余数之和的编程

#includeintmain(){inta,b,c;scanf("%d%d%d",&a,&b,&c);if(a+b>c&&b==c)printf("1");elseprintf("0");}

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(){inta,b;scanf("%d%d",&a,&b);printf("%d",a-b);return0;}

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

//基础题看看吧#include#includeboolprim(intn){if(n

#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

#includeintmain(){intx,y;while(scanf("%d%d",&x,&y)==2&&(x!=0||y!=0)){\x09printf("%d\n",x+y);}return0

#includevoidmain(){inta,b,sum;scanf("%d,%d",&a,&b);sum=a+b;printf("thesumis%d",sum);}

#include"stdio.h"#include"math.h"main(){inta,b,i,j,k,t,n=0;printf("请输入两个整数:");scanf("%d%d",&a,&b);if

voidfun(int*a,int*b){intc=0;c=*a;*a=*b;*b=c;return;}

#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

programab;varA,B,i,s:integer;beginwrite('请输入两个自然数’）；readln(A,B);i:=1;s:=m*i;whilesmodBOdobegini:=i+1

a²+b²+c²+4≤ab+3b+2ca²-ab+b²/4+3b²/4-3b+3+c²-2c+1≤0(a-b/2)²+3

#includeusingnamespacestd;intmain(){intx,y,z;while(1){coutx>>y>>z;cout

Subtest()Dimarr(1To3)AsIntegerDimiAsIntegerDimaAsInteger,bAsInteger,cAsIntegerDimxAsStringFori=1To3a

算法程序：#includevoidmain(){inta,b;printf("inputa,b:(a>b)\n");scanf("%d,%d",&a,&b);if(a%b==0)printf("OK\

a/b等到的是商,a%b求得的是余数.由于是整数,所以a/b得到的是整数,余数部分被省略,需要另求!

#includeusingnamespacestd;voidmain(){floata,b,c,d,max;couta>>b>>c>>d;max=(a>b)?a:b;max=(max>c)?max:c

#includeintGcd(intM,intN){intRem;while(N>0){Rem=M%N;M=N;N=Rem;}returnM;}voidmain(){inta,b