1065 A+B and C (64bit) （20 分）

1065 A+B and C (64bit) （20 分）

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

Sample Output:

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

分析：这个题有坑点，涉及到计算机组成原理的知识，数字的范围是[−2^​63​​,2​^63​​]， 而long long 的范围是[-2^63，2^63)，两个数相加会溢出，符号位会取反。

 /**
 * Copyright(c)
 * All rights reserved.
 * Author : Mered1th
 * Date : 2019-02-23-19.49.08
 * Description : A1065
 */
 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<string>
 #include<unordered_set>
 #include<map>
 #include<vector>
 #include<set>
 using namespace std;

 int main(){
 #ifdef ONLINE_JUDGE
 #else
     freopen("1.txt", "r", stdin);
 #endif
     ;
     scanf("%d",&n);
     ;i<=n;i++){
         long long a,b,c;
         scanf("%lld%lld%lld",&a,&b,&c);
         bool flag;
         long long res=a+b;
         &&b>&&res<) flag=true;
         &&b<&&res>=) flag=false;
         else if(res>c) flag=true;
         else flag=false;
         if(flag==true) printf("Case #%d: true\n",i);
         else printf("Case #%d: false\n",i);
     }

     ;
 }