AtCoder 2333 Scc Puzzle （水题）

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这是一篇关于AtCoder编程挑战中Scc Puzzle问题的解析，讲述了如何使用两个C形块组合成一个S形块，以创建尽可能多的Scc组合。题目给出输入输出格式，并提供了一个示例解释如何计算最大数量的Scc组。解决方案提到，先将S与C配对，然后计算剩余C的数量除以4加上剩余S的数量即为答案。

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题目链接：http://abc055.contest.atcoder.jp/tasks/arc069_a?lang=en

C - Scc Puzzle

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

Snuke loves puzzles.

Today, he is working on a puzzle using S- and c-shaped pieces. In this puzzle, you can combine two c-shaped pieces into one S-shaped piece, as shown in the figure below:

Snuke decided to create as many Scc groups as possible by putting together one S-shaped piece and two c-shaped pieces.

Find the maximum number of Scc groups that can be created when Snuke has N S-shaped pieces and M c-shaped pieces.

Constraints

1≤N,M≤1012

Input

The input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

Copy

1 6

Sample Output 1

Copy

Two Scc groups can be created as follows:

Combine two c-shaped pieces into one S-shaped piece
Create two Scc groups, each from one S-shaped piece and two c-shaped pieces

Sample Input 2

Copy

12345 678901

Sample Output 2

Copy

Submit

题意：输入n个S， m个C，一个S可以由2个C组成，求一共有多少个Scc

解析：先把S与C配对，剩下的C/4 + S的个数就是答案

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
#define N 1009
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long LL;

LL fun(LL n, LL m)
{
    if((m / 2) < n) return  m / 2;
    LL ans = n;
    m = m - n * 2;
    ans = ans + m / 4;
    return ans;
}


int main()
{
    LL n, m;
    scanf("%lld%lld", &n, &m);
    LL ans = fun(n, m);
    printf("%lld\n", ans);
    return 0;
}