Ternary Calculation

G-TernaryCalculation详解

最新推荐文章于 2019-06-30 01:30:49 发布

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本文详细介绍了G-TernaryCalculation问题的解决方法，包括输入格式、运算规则及输出示例。通过实例演示了如何进行三元运算并计算最终结果。

G - Ternary Calculation

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3782

Description

Complete the ternary calculation.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is a string in the form of "number₁operator_anumber₂operator_bnumber₃". Each operator will be one of {'+', '-' , '*', '/', '%'}, and each number will be an integer in [1, 1000].

Output

For each test case, output the answer.

Sample Input

Sample Output

Note

The calculation "A % B" means taking the remainder of A divided by B, and "A / B" means taking the quotient.

#include<stdio.h>
#include<string.h>
int main()
{
	int a1[4],n;
	char c[3];
	while(~scanf("%d",&n))
	{while(n--)
	{
	
		int s,m,o,k,l,i;
		s=m=o=k=0;
		scanf("%d %c %d %c %d",&a1[0],&c[0],&a1[1],&c[1],&a1[2]);
		if((c[0]=='+'||c[0]=='-')&&(c[1]=='*'||c[1]=='/'||c[1]=='%'))	
		{
			if(c[1]=='*')
		     s=a1[1]*a1[2];
			else if(c[1]=='/')
			s=a1[1]/a1[2];
			else s=a1[1]%a1[2];
			if(c[0]=='+')
			s+=a1[0];
			else
			s=a1[0]-s;
			printf("%d\n",s);
		}
		else	
		{
			if(c[0]=='+')
			s=a1[1]+a1[0];
			else if(c[0]=='-')
			s=a1[0]-a1[1];
			else if(c[0]=='*')
			s=a1[0]*a1[1];
			else if(c[0]=='/')
			s=a1[0]/a1[1];
			else s=a1[0]%a1[1];
			if(c[1]=='+')
			s+=a1[2];
			else if(c[1]=='-')
			s-=a1[2];
			else if(c[1]=='*')
			s*=a1[2];
			else if(c[1]=='/')
			s/=a1[2];
			else s=s%a1[2];
			printf("%d\n",s);
		}
	}
}
}