1065 A+B and C (64bit) (20 分)（模拟）

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：

因为A、B的大小为[-2^63, 2^63]，用long long 存储A和B的值，以及他们相加的值sum：

如果A > 0, B < 0 或者 A < 0, B > 0，sum是不可能溢出的

如果A > 0, B > 0，sum可能会溢出，sum范围理应为(0, 2^64 – 2]，溢出得到的结果应该是[-2^63, -2]是个负数，所以sum < 0时候说明溢出了

如果A < 0, B < 0，sum可能会溢出，同理，sum溢出后结果是大于0的，所以sum > 0 说明溢出了

原文链接：https://blog.csdn.net/liuchuo/article/details/52109211

题解

注意用用cin读取数据最后一个样例报错~

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        long long int a,b,c;
        scanf("%lld %lld %lld", &a, &b, &c);
        long long int sum=a+b;
        if(a>0&&b>0&&sum<0){
            printf("Case #%d: true\n",i);
        }else if(a<0&&b<0&&sum>=0)
            printf("Case #%d: false\n",i);
        else if(sum>c)
            printf("Case #%d: true\n",i);
        else
            printf("Case #%d: false\n",i);
    }

    return 0;
}