AtCoder题解 —— AtCoder Beginner Contest 188 —— A - Three-Point Shot

题目相关

题目链接

AtCoder Beginner Contest 188 A 题，https://atcoder.jp/contests/abc188/tasks/abc188_a。

Problem Statement

A basketball game is being played, and the score is now X-Y. Here, it is guaranteed that $X\ne Y$.
Can the team which is behind turn the tables with a successful three-point goal?
In other words, if the team which is behind earns three points, will its score become strictly greater than that of the other team?

Input

Input is given from Standard Input in the following format:

X Y

Output

If the team which is behind can turn the tables with a successful three-point goal, print Yes; otherwise, print No.

Sample 1

Sample Input 1

3 5

Sample Output 1

Yes

Explaination

The team with $3$ points is behind.
After a successful $3$-point goal, it will have $6$ points, which is greater than that of the other team - $5$.

Sample 2

Sample Input 2

16 2

Sample Output 2

No

Explaination

The gap is too much. The team which is behind cannot overtake the other by getting $3$ points.

Sample 3

Sample Input 3

12 15

Sample Output 3

No

Explaination

A $3$-point goal will tie the score but not turn the tables.

Constraints

• $0\le X\le 100$
• $0\le Y\le 100$
• $X\ne Y$
• $X$ and $Y$ are integers.

题解报告

题目翻译

有两个队进行篮球比赛，对应的比分是 $X$ - $Y$，保证 $X\ne Y$。问落后的队伍能否通过一次 $3$ 分球获得胜利。

题目分析

一个简单的数学题。
直接的思路就是判断 $X$ 和 $Y$ 的大小，假设 $Y$ 较小，如果 $Y+3>X$，那么可以获胜，否则不能获胜。
经过观察，我们发现只需要两队比分差再 $3$ 分以内，落后方才能通过一个 $3$ 分球获胜。也就是说当满足 $|X-Y|<3$ 的时候，才可能出现这样的情况。

数据范围分析

ABC 的 A 题，数据范围都很小。根据题目可知，int 足够。

AC 代码

//https://atcoder.jp/contests/abc188/tasks/abc188_a
//A - Three-Point Shot
#include <bits/stdc++.h>

using namespace std;

//如果提交到OJ，不要定义 __LOCAL
//#define __LOCAL

int main() {
    
    
#ifndef __LOCAL
    //这部分代码需要提交到OJ，本地调试不使用
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#endif
    int x,y;
    cin>>x>>y;

    if (abs(x-y)<3) {
    
    
        cout<<"Yes\n";
    } else {
    
    
        cout<<"No\n";
    }

#ifdef __LOCAL
    //这部分代码不需要提交到OJ，本地调试使用
    system("pause");
#endif
    return 0;
}

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时间复杂度

O(1)。

空间复杂度

O(1)。