ZOJ3782-Ternary Calculation
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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 "number1 operatora number2 operatorb number3". 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
51 + 2 * 31 - 8 / 31 + 2 - 37 * 8 / 55 - 8 % 3
Sample Output
7-10113
Note
The calculation "A % B" means taking the remainder of A divided by B, and "A / B" means taking the quotient.
Author: ZHOU, Yuchen
Source: The 11th Zhejiang Provincial Collegiate Programming Contest
题意:给你一串算式,计算出答案
#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <map>#include <vector>#include <queue>#include <bitset>#include <stack>using namespace std;int main(){ char ch1,ch2; int a,b,c; int t; scanf("%d",&t); while(t--) { scanf("%d %c %d %c %d",&a,&ch1,&b,&ch2,&c); if(ch1=='+'||ch1=='-') { int ans; if(ch2=='+') ans=b+c; else if(ch2=='-') ans=b-c; else if(ch2=='*') ans=b*c; else if(ch2=='/') ans=b/c; else ans=b%c; if(ch1=='+') printf("%d\n",ans+a); else { if(ch2=='-') printf("%d\n",a-b-c); else if(ch2=='+') printf("%d\n",a-b+c); else printf("%d\n",a-ans); } } else { int ans; if(ch1=='*') ans=a*b; else if(ch1=='/') ans=a/b; else if(ch1=='%') ans=a%b; if(ch2=='+') printf("%d\n",ans+c); else if(ch2=='-') printf("%d\n",ans-c); else if(ch2=='*') printf("%d\n",ans*c); else if(ch2=='/') printf("%d\n",ans/c); else printf("%d\n",ans%c); } } return 0;}
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