1065 A+B and C (64bit) （20 分）

Given three integers A, B and C in [−2
​63
​​ ,2
​63
​​ ], 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

判断a+b是否大于c。这里需要注意的是a+b可能溢出，如果a大于0且b大于0，但相加得到的却是小于等于0的说明
是正溢出，这时肯定比c大（因为c肯定在long long的范围内）。如果a小于0且b小于0，但相加得到的却是大于
等于0的说明是负溢出，这是肯定比c小。其他情况就和平常计算一样。这里还要注意a+b要赋值给一个变量再和c
比较。
--------------------- 
原文：https://blog.youkuaiyun.com/kakitgogogo/article/details/51954542 

AC代码：
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
    int T;
    int t = 0;
    long long a,b,c;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld%lld",&a,&b,&c);
        printf("Case #%d: ",++t);
        long long res = 0;
        res = a+b;
        if(a>0&&b>0&&res<=0)
            printf("true\n");
        else if(a<0&&b<0&&res>=0)
            printf("false\n");
        else if(res>c)
            printf("true\n");
        else printf("false\n");
    }
    return 0;
}