1065 A+B and C (64bit) (20分)(水)

Given three integers A, B and C in [−], 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 (≤). 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

题目分析：利用long long 存储数据 但是要注意对于num的判断 因为long long 最大存储范围是从负2的63次 到 正2的63次减一
如果这个和大于或小于这个范围 就要先判断再输出
对于这道题来说
数的范围是 $[-2^63,2^63]$ 而longlong的范围是$[-2^63,2^63-1]$
若两个数的和过大到区间$[2^63,2^64-2] 那么会溢出改变到区间 [-2^63,-2](对于非浮点数来说上界加一会变为加一后的值取负 故(2^63-1+1)会变为-2^63 -2是根据公式 (2^64-2)%2^64=-2求得$
同理 若过小到区间$[-2^64,-2^63-1] 会溢出改变到区间[0,2^63-1]$

#define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
string A, B, C;

int main()
{
    int N;
    cin >> N;
    for (int i = 1; i <= N; i++)
    {
        long long A, B, C;
        cin >> A >> B >> C;
        long long num = A + B;
        if (A > 0 && B > 0 && num < 0)
            printf("Case #%d: true
", i);
        else if (A < 0 && B < 0 && num >= 0)
            printf("Case #%d: false
", i);
        else if(num > C)
            printf("Case #%d: true
", i);
        else
            printf("Case #%d: false
", i);
    }
}