1065 A+B and C (64bit)_1065 a+b and c (64bit)-优快云博客

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

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

不得不说，PAT是一个非典型的编程测试。这个题居然和计算机ALU实际进行加法的原理扯上了关系。

对于long long int，可以表达的范围正好是 $[-2^{63},2^{63}]$ ，虽然在两个数同号的时候，它们的和可能不能表示，但是我们可以根据和的正负分别判断出现正溢出或者负溢出。（参考《汇编程序语言设计》）

1. 当A>0，B>0，A+B<0时为正溢出，此时输出true。

2. 当A<0，B<0，A+B>=0时，为负溢出，此时输出false。

#include<iostream>
#define LL long long
using namespace std;
bool judge(LL a,LL b,LL c){
	LL sum = a+b;
	if(a>0&&b>0&&sum<0) return true;//正溢出
	else if(a<0&&b<0&&sum>=0) return false;//负溢出
	else if(sum>c) return true;//没有发生溢出
	else if(sum<=c) return false;//同上 
}
int main(void){
	int T;
	cin>>T;
	LL a,b,c;
	for(int i=0;i<T;i++){
		cin>>a>>b>>c;
		cout<<"Case #"<<i+1<<": ";
		if(judge(a,b,c)) cout<<"true"<<endl;
		else cout<<"false"<<endl;
	}
	return 0;
}

1. int和LL的表示范围：https://blog.youkuaiyun.com/gotowu/article/details/51203920

2. 参考链接：https://blog.youkuaiyun.com/richenyunqi/article/details/80110078