ZJU-PAT 1065. A+B and C (64bit) (20)

判断整数加法溢出

最新推荐文章于 2022-09-24 11:48:38 发布

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本文介绍了一个程序设计问题，即如何判断三个给定整数A、B和C是否满足A+B>C的条件，特别关注了整数溢出的情况，并提供了一种使用C++实现的解决方案。

Given three integers A, B and C in [-2⁶³, 2⁶³], you are supposed to tell whether A+B > C.

Input Specification:

The first line of the input gives the positive number of test cases, T (<=10). Then T test cases follow, each consists of a single line containing three integers A, B and C, separated by single spaces.

Output Specification:

For each test case, output in one line "Case #X: true" if A+B>C, or "Case #X: false" otherwise, where X is the case number (starting from 1).

Sample Input:

3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: false
Case #2: true
Case #3: false

分析：
1.考虑相加溢出的情况
(1)  a > 0 && b>0 && a+b <0  , 因 a+b > c，故true
(2)  a < 0 && b<0 && a+b >0 , 因 a+b < c，故false
2. PAT上不支持 __int64数据,采用long long型数据

#pragma warning (disable:4786)
#include<iostream>
#include<string>
#include<cstdio>
#include<algorithm>
#include<map>
using namespace std;

int main()
{
    int T;
    cin>>T;
    for(int i=1;i<=T;i++)
    {
        long long int  a,b,c;
        cin>>a>>b>>c;

        cout<<"Case #"<<i+1<<": ";
        
        bool flag;
        long long ans =a+b;
        if(a>0 && b>0 && ans<=0)
            flag=true;
        else if(a<0 && b <0 && ans >= 0)
            flag=false;
        else
            flag = (ans > c);
        if(flag)
            cout<<"true"<<endl;
        else
            cout<<"false"<<endl;
    }
    return 0;
}