第二次组队赛（xiao）codeforces1070

最新推荐文章于 2023-04-23 17:03:33 发布

ZCY19990813

最新推荐文章于 2023-04-23 17:03:33 发布

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分类专栏：组队赛

本文链接：https://blog.youkuaiyun.com/ZCY19990813/article/details/83239271

本文包含了四道编程竞赛题目：关于视频是否能均分到帖子中、如何高效处理垃圾、选举辩论赛观众选择策略以最大化影响力，以及BerOS文件系统的建议方案。每道题目提供了详细的理解和解题思路。

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K. Video Posts

Input

The first line contains two integers n and k (1≤k≤n≤105). The next line contains n positive integer numbers a1,a2,…,an (1≤ai≤104), where ai is the duration of the i-th video.

Output

If solution exists, print "Yes" in the first line. Print k positive integers s1,s2,…,sk (s1+s2+⋯+sk=n) in the second line. The total duration of videos in each post should be the same. It can be easily proven that the answer is unique (if it exists).

If there is no solution, print a single line "No".

Examples

Input

6 3
3 3 1 4 1 6

Output

Yes
2 3 1

Input

3 3
1 1 1

Output

Yes
1 1 1

Input

3 3
1 1 2

Output

No

Input

3 1
1 10 100

Output

Yes
3

题意：就是看能否均分，是的话就Yes,然后输出可以由几个数组成，不是就No

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<math.h>
#include<map>
#include<string>
using namespace std;
typedef long long ll;
map<int,int>mp;
map<int,int>::iterator it;
ll c[100010],a[101010];
int main()
{
    ll n,k,i,j,s=0,ss;
    cin>>n>>k;
    for(i=0; i<n; i++)
    {
        cin>>a[i];
        s+=a[i];
    }
    if(s%k)
        printf("No\n");
    else
    {
        ss=s/k;
        ll e=0,ans=0,p=0;
        memset(c,0,sizeof(c));
        for(i=0; i<n; i++)
        {
            ans+=a[i];
            p++;
            if(ans==ss)
            {
                c[e++]=p;
                ans=0;
                p=0;
            }
        }
        //cout<<e<<endl;
        if(e==k)
        {
            cout<<"Yes"<<endl;
            cout<<c[0];
            for(i=1; i<e; i++)
                cout<<" "<<c[i];
            cout<<endl;
        }
        else
            cout<<"No"<<endl;
    }

    return 0;
}

D. Garbage Disposal

Examples

Input