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

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本文介绍了一种解决特定整数运算问题的方法：给定三个整数A、B和C，判断A+B是否大于C。文章通过分析整数的取值范围及加法运算的特点，提出了一种高效的解决方案，并附带了实现该算法的C++代码。

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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

题解

考虑到a, b, c的取值范围，这题并不需要模拟大数的加法。

有符号long long的范围是 - $2^{63}$ ~ $2^{63}$ - 1

a + b只有同为正数或同为负数才有可能溢出，溢出必然变号，也肯定大于（小于）c。

#include <bits/stdc++.h>
using namespace std;

long long a, b, c;
int n;

int main(){

    cin >> n;
    for(int i = 1; i <= n; ++i){
        cin >> a >> b >> c;
        cout << "Case #" << i << ": ";
        long long t = a + b;
        if(a > 0 && b > 0 && t < 0) cout << "true" << endl;
        else if(a < 0 && b < 0 && t >= 0) cout << "false" << endl;
        else if(t > c) cout << "true" << endl;
        else cout << "false" << endl;
    }


    return 0;
}