CodeForces 839B_coderforces 839b-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_44751481/article/details/89668321

在本篇博客中，我们探讨了一个基于贪心算法的游戏问题——如何在限制条件下合理安排不同队伍的士兵乘坐飞机座位，以避免来自不同队伍的士兵坐在相邻座位上。通过对座位的巧妙划分和对队伍人数的合理分配，我们成功解决了这一挑战。

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B. Game of the Rows

Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.

在这里插入图片描述 A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.

Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.

Input

The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.

The second line contains k integers a1, a2, a3, …, ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.

It is guaranteed that a1 + a2 + … + ak ≤ 8·n.

Output

If we can place the soldiers in the airplane print “YES” (without quotes). Otherwise print “NO” (without quotes).

You can choose the case (lower or upper) for each letter arbitrary.

Examples

Input
2 2
5 8
Output
YES

Input
1 2
7 1
Output
NO

Input
1 2
4 4
Output
YES

Input
1 4
2 2 1 2
Output
YES

Note

In the first sample, Daenerys can place the soldiers like in the figure below:
在这里插入图片描述

In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.

In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.

In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).

思路

贪心，中间四个位置是连在一起的，但是可以分成1+2的形式，两个四连座，可以分给三个2人队伍坐。因为四个连座比较难处理，我刚开始的想法是，输入一次就先安排上了，但是后面可能我分的有点问题，出现了一点bug。所以我后来决定先录入，再循环处理不同情况，优先处理能坐4人的，然后处理能坐2人的，最后处理剩下的1人或者还有别的队伍没有安排。

代码

#include<stdio.h>
int main() {
	int n,k;
	while(scanf("%d %d",&n,&k) != EOF) {
		int shuangzuo = n;
		int danzuo = n * 2;
		int a[10000+5];
		int t = 0;
		for(int i=0;i<k;i++) {
			scanf("%d",&a[i]);
		}
		for(int i=0;i<k;i++) {
			if(a[i] >= 4) {
				if(shuangzuo < a[i] / 4) {
					t = shuangzuo;
				}
				else {
					t = a[i] / 4;
				}
				a[i] -= 4 * t;
				shuangzuo -= t;
			}
		}
		danzuo += shuangzuo;
		for(int i=0;i<k;i++) {
			if(a[i] >= 2) {
				if(danzuo < a[i] / 2) {
					t = danzuo;
				}
				else {
					t = a[i] / 2;
				}
				a[i] -= 2 * t;
				danzuo -= t;
			}
		}
		danzuo += shuangzuo;
		for(int i=0;i<k;i++) {
			if(a[i] >= 1) {
				danzuo -= a[i];
			}
		}
		if(shuangzuo >=0 && danzuo >= 0) {
			printf("YES\n");
		}
		else {
			printf("NO\n");
		}
	}
}