1065 A+B and C (64bit)

1065 A+B and C (64bit) （20 分）

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 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

a,b,c,sum(=a+b)均定义为long long,范围为[-2⁶³,2⁶³-1],当a，b同号时相加可能存在溢出，当a,b均为正数时，a+b的范围是(0,2⁶⁴-2],溢出的范围是[2⁶³,2⁶⁴-2],在long long范围内表示为[2^-63,-2],所以溢出时，sum<0,2⁶³==2^-63,2⁶⁴==0,同理当a,b均为负数时，a+b的范围是[-2⁶⁴,0),在[-2⁶⁴,-2^-63-1]范围内会溢出，这个范围在long long中表示为[0,2⁶³-1],所以溢出时，sum>=0,当sum没有溢出时，可以按照常规判断。
注：关于整形数的表示范围的推演可以参考博客^[1]或者阅读计算机组成原理相关书籍

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

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1. https://www.cnblogs.com/luojialin/p/4764745.html ↩