#include <stdio.h>
int main ()
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
printf("%d+%d+%d=%d\n",a,b,c,a+b+c);
printf("%d-%d-%d=%d\n",a,b,c,a-b-c);
printf("%d*%d*%d=%d\n",a,b,c,a*b*c);
printf("%d/%d/%d=%d\n",a,b,c,a/b/c);
printf("%d%%%d%%%d=%d\n",a,b,c,a%b%c);
return 0;
int main ()
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
printf("%d+%d+%d=%d\n",a,b,c,a+b+c);
printf("%d-%d-%d=%d\n",a,b,c,a-b-c);
printf("%d*%d*%d=%d\n",a,b,c,a*b*c);
printf("%d/%d/%d=%d\n",a,b,c,a/b/c);
printf("%d%%%d%%%d=%d\n",a,b,c,a%b%c);
return 0;
}
得到
1 2 3
1+2+3=6
1-2-3=-4
1*2*3=6
1/2/3=0
1%2%3=1
本例为3个整数的运算问题,注意%的实现