A1065. A+B and C (64bit)

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

#include<cstdio>
#include<iostream>
using namespace std;
int main(){
    int T;
    long long a, b, c, test;
    scanf("%d", &T);
    for(int i = 0; i < T; i++){
        scanf("%lld%lld%lld", &a, &b, &c);
        test = a + b;
        if(a >0 && b > 0 && test < 0){
            printf("Case #%d: true
", i + 1);
        }else if(a < 0 && b < 0 && test >= 0){
            printf("Case #%d: false
", i + 1);
        }else{
            if(test > c)
                printf("Case #%d: true
", i + 1);
            else
                printf("Case #%d: false
", i + 1);
        }
    }
    return 0;
}

总结：1、首先输入的范围在[-2^63, 2^63], 则只能用long long 8个byte存储。当a与b过大时，会产生溢出，需要考虑正数+正数=负数及负数+负数=正数的情况。

　　　2、A + B的结果不能在条件语句中直接与C比较，必须先存储于long long变量中。