#1059. A+B and C (64bit)【模拟】

原题链接

Given three integers A , B A, B A,B and C C C in $(-2^{63}, 2^{63}), you are supposed to tell whether A + B > C A + B > C A+B>C.

The first line of the input gives the positive number of test cases, T ( ≤ 10 ) T(leq 10) T(≤10). Then T T T test cases follow, each consists of a single line containing three integers A , B A, B A,B and C C C, separated by single spaces.

For each test case, output in one line Case #X: true if A + B > C A + B > C A+B>C, or Case #X: false otherwise, where X X X is the case number (starting from 1).

3
1 2 3
2 3 4
9223372036854775807 -9223372036854775808 0

Case #1: false
Case #2: true
Case #3: false

本题涉及到 64 64 64 位整数相加，相加过程会爆 long long，因此我们将所有情况列出来，进行判断即可。

1. a > 0 , b > 0 , a + b < 0 a > 0, b > 0, a + b < 0 a>0,b>0,a+b<0，正向溢出，此时 a + b a + b a+b 一定大于 c c c。
2. a < 0 , b < 0 , a + b ≥ 0 a < 0, b < 0, a + b geq 0 a<0,b<0,a+b≥0，反向溢出，此时 a + b a + b a+b 一定小于 c c c。
3. 均无溢出，此时直接判断 a + b a + b a+b 是否大于 c c c 即可。

#include 
#include 
#include 
#include 

using namespace std;

typedef long long LL;

bool check(LL a, LL b, LL c)
{
    LL res = a + b;
    if (a > 0 && b > 0 && res < 0) return true;
    if (res > c) return true;
    return false;
}

int main()
{
    int T;
    cin >> T;
    for (int i = 1; i <= T; i ++ )
    {
        LL a, b, c;
        cin >> a >> b >> c;
        if (check(a, b, c)) printf("Case #%d: truen", i);
        else printf("Case #%d: falsen", i);
    }
    return 0;
}