1065 A+B and C (64bit) (20分)

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本文介绍了一个关于64位整数溢出的编程问题，通过将大数分解成两部分的方法来解决A+B>C的判断问题，提供了一段C++代码实现。

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 Ttest 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

题意：判断a + b 是否大于 c，这里牵扯一个越界问题，64位（long long int）整数来说，补码的范围为[-2^63,2^63)，最后这个数是取不到的，但是这个题好像没有出这个点难为人，那就很简单了，把一个大数分解成两部分不就ok了吗，直接莽就行～

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int main(){
    
    int t;
    cin >> t;
    long long int temp =  pow(2, 60);
    for(int i = 1; i <= t; i++){

        long long int a, b, c;
        scanf("%lld %lld %lld", &a, &b, &c);
        
        long long int pre_a = a / temp;
        long long int back_a = a % temp;
        long long int pre_b = b / temp;
        long long int back_b = b % temp;
        long long int pre_c = c / temp;
        long long int back_c = c % temp;
        if(pre_a + pre_b > pre_c){
            printf("Case #%d: true\n", i);
        }else if (pre_a + pre_b == pre_c && back_a + back_b > back_c)
        {
            printf("Case #%d: true\n", i);
        }else
        {
            printf("Case #%d: false\n", i);
        }
        
        

    }
    
    



    return 0;
}