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:

31 2 32 3 49223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: falseCase #2: trueCase #3: false

加法溢出。只有当两个正数相加或者两个复数相加才可能溢出（相同类型的两个数）。两个整数相加为复数则溢出。两个复数相加为非负数则溢出。

#include <iostream>using namespace std;int main(){long long a,b,c;int n;cin>>n;for(int i=0;i<n;i++){cin>>a>>b>>c;long long ans=a+b;if((a > 0 && b >0) && ans < 0)//上益printf("Case #%d: true\n",i+1);else if((a<0 && b< 0) && ans>= 0)//下益printf("Case #%d: false\n",i+1);else{if(a+b>c)printf("Case #%d: true\n",i+1);elseprintf("Case #%d: false\n",i+1);}}return 0;}