1065 A+B and C (64bit) (20)

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&gtC, 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

题目大意：判断两数之和是否大于第三个数；
思路：因为输入的数的范围是[-2^63, 2^63] ，且c++中最大整数保存范围不超过2^64，则有相加的时候可能出现溢出；开始想的是把输入的数转换为2进制补码运算，还是比较麻烦，但是是可行的。
；参考了别人的做法；对于溢出的情况不需要算出起正确的值。因为出只有两种情况整数相加，和负数相加。前者之和肯定比第三者大， 后者之后肯定比第三者小。所以当符号相等的时候，只需要判断和是否溢出就能判断三者的大小关系；其余的按照正常的运算来比较
参考地址：https://www.liuchuo.net/archives/2023

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