1065. A+B and C (64bit) (20)

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

---

 思路：根据是否溢出来做 ，但是好郁闷    arr[i].a+arr[i].b作为整体时就出错！ long long total = arr[i].a+arr[i].b;  用total代替就OK' 真心搞不懂啊！！

AC代码：

#include <iostream>
using namespace std;
int N;
struct Node{
    long long a;
    long long b;
    long long c;
};
Node arr[11];
int main()
{
    cin>>N;
    for(int i=0;i<N;i++){
        cin>>arr[i].a>>arr[i].b>>arr[i].c;
        long long total = arr[i].a+arr[i].b;
        if(arr[i].a<0 && arr[i].b<0 && total>=0){   //发生溢出
            cout<<"Case #"<<i+1<<": "<<"false"<<endl;
        }else if(arr[i].a>0 && arr[i].b>0 && total<=0){
            cout<<"Case #"<<i+1<<": "<<"true"<<endl;
        }else{
            if(total>arr[i].c){
                cout<<"Case #"<<i+1<<": "<<"true"<<endl;
            }else{
                cout<<"Case #"<<i+1<<": "<<"false"<<endl;
            }
        }
    }
    return 0;
}