PAT甲级——（1065） A+B and C(64bit)

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

题意：三个数A、B、C位于 [−2^​63​​,2 ^63​​]，判断A+B是否大于C
思路：用long long来存ABC，关键要判断A+B是否有溢出，要知道，如果两个正数相加溢出的话，相加的值中超出范围的值会从−2^{​63​​数过来。即值会在[−2}​63​​,-2​​]之间（是-2的原因是long long 范围内负数比整数多表示1个）。如果两个负数相加溢出的话，相加的值中超出范围的值会从2的63次方减1​​数过来。即值会在[2^​63-1​​,0​​]之间。

#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 sum;
		sum = a + b;
		if (a > 0 && b > 0 && sum < 0) {
			printf("Case #%d: true\n", i + 1);
		}
		else if(a < 0 && b < 0 && sum>=) {
			printf("Case #%d: false\n", i + 1);

		}else if(sum>c){
			printf("Case #%d: true\n", i + 1);
		}
		else {
			printf("Case #%d: false\n", i + 1);
		}
	}
	return 0;
}