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  • PAT1065: A+B and C (64bit)

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

    时间限制
    100 ms
    内存限制
    65536 kB
    代码长度限制
    16000 B
    判题程序
    Standard
    作者
    HOU, Qiming

    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

    思路
    注意溢出的情况就行,其他情况正常比较,溢出分两种情况:
    1. A > 0,B > 0, Sum = A + B 正溢出,溢出后的值Sum <= 0;
    2. A < 0,B < 0, Sum = A + B 负溢出,溢出后的值Sum >= 0;

    代码
    #include<iostream>
    #include<vector>
    using namespace std;
    int main()
    {
        int T;
        while(cin >> T)
        {
            long long a,b,c;
            for(int i = 1; i <= T; i++)
            {
                cin >> a >> b >> c;
                long long sum = a + b;
    
                if(a < 0 && b < 0 && sum >= 0) //最小负数相加溢出
                    cout << "Case #" << i <<": false" << endl;
                else if(a > 0 && b > 0 && sum <= 0) //最大正数相加溢出
                    cout << "Case #" << i <<": true" << endl;
                else if(sum > c)
                    cout << "Case #" << i <<": true" << endl;
                else
                    cout << "Case #" << i <<": false" << endl;
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/0kk470/p/7631636.html
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