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

这道题刚看到还准备用大数加法起笔呢，后来发现没那么麻烦，几个条件判断一下就好了。 新知道两个点，两个正数相加，溢出的和是负的，两个负数相加溢出的话，和是正的！

#include<cstdio>
#include <cstring>
#include<algorithm>
#include <iostream>
#include <cmath>
#define ll long long
using namespace std;

int main()
{
    ll a,b,c,sum;
    int t;
    scanf("%d",&t);
    for(int i = 1; i <= t;i++)
    {
        sum = 0;
        scanf("%lld%lld%lld",&a,&b,&c);
        sum = a + b;
        //cout<<sum<<endl;
        if(a > 0 && b > 0 && sum < 0)//1
            printf("Case #%d: true\n",i);
        else if(a < 0 && b < 0 && sum >= 0)//2.注意别忘等于号！
            printf("Case #%d: false\n",i);
        else if(sum > c)//3.条件2含在条件3里面，所以要先剔除条件2的情况
            printf("Case #%d: true\n",i);
        else
            printf("Case #%d: false\n",i);//4
    }
    
    return 0;
}