【1065】 A+B and C (64bit) 溢出检测

2024-04-05 06:18

文章标签 检测溢出 1065 64bit

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

时间限制

100 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

HOU, Qiming

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

题意：

给出三个整数，判断前两个的和是不是大于第三个。

分析：

考察溢出的检测问题, 当A和B同时为正整数或者同时为负整数的时候要小心题目的和可能溢出, 所以不能简单的return A+B > C, 要加溢出的判断代码。

#define LLONG_MAX 9223372036854775807i64 /* maximum signed long long int value */
#define LLONG_MIN (-9223372036854775807i64 - 1) /* minimum signed long long int value */

代码：

#include <iostream>
#include <fstream>
#include <climits>
using namespace std;ifstream fin("in.txt");
#define cin finbool Compare(const long long& a,const long long& b,const long long& c){if(a>0 && b>0){if(a > LLONG_MAX - b)return true;}if(a<0 && b<0){if(a < LLONG_MIN -b)return false;}return a+b > c;
}int main()
{int n;cin>>n;long long a,b,c;int i;for(i=0;i<n;i++){cin>>a>>b>>c;cout<<"Case #"<<i+1<<": ";if(Compare(a,b,c)){cout<<"true"<<endl;}elsecout<<"false"<<endl;}system("PAUSE");return 0;
}

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