1 solutions
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0
C :
#include <stdio.h> int main() { double num1,num2; scanf("%lf %lf",&num1,&num2); printf("%.0lf+%.0lf=%.0lf\n",num1,num2,num1 + num2); printf("%.0lf-%.0lf=%.0lf\n",num1,num2,num1 - num2); printf("%.0lf*%.0lf=%.0lf\n",num1,num2,num1 * num2); printf("%.0lf/%.0lf=%.2lf\n",num1,num2,num1 / num2); printf("%.0lf div %.0lf=%d\n",num1,num2,(int)num1 / (int)num2); printf("%.0lf mod %.0lf=%d\n",num1,num2,(int)num1 % (int)num2); return 0; }C++ :
#include<cstdio> using namespace std; int main() { int a,b; scanf("%d%d",&a,&b); printf("%d+%d=%d\n",a,b,a+b); printf("%d-%d=%d\n",a,b,a-b); printf("%d*%d=%d\n",a,b,a*b); printf("%d/%d=%.2lf\n",a,b,a/b*1.0); printf("%d div %d=%d\n",a,b,a/b); printf("%d mod %d=%d\n",a,b,a%b); return 0; }
- 1
Information
- ID
- 18543
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By