hdu5355 思维+爆搜

Posted 2020-10-25 zhchoutai

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pid=5355">http://acm.hdu.edu.cn/showproblem.php?pid=5355

Problem Description

There are?m?soda and today is their birthday. The?1-st soda has prepared?n?cakes with size?1,2,…,n. Now?1-st soda wants to divide the cakes into?m?parts so that the total size of each part is equal.?

Note that you cannot divide a whole cake into small pieces that is each cake must be complete in the?m?parts. Each cake must belong to exact one of?m?parts.

?

Input

There are multiple test cases. The first line of input contains an integer?T, indicating the number of test cases. For each test case:

The first contains two integers?n?and?m?(1≤n≤105,2≤m≤10), the number of cakes and the number of soda.
It is guaranteed that the total number of soda in the input doesn’t exceed 1000000. The number of test cases in the input doesn’t exceed 1000.

?

Output

For each test case, output "YES" (without the quotes) if it is possible, otherwise output "NO" in the first line.

If it is possible, then output?m?lines denoting the?m?parts. The first number?si?of?i-th line is the number of cakes in?i-th part. Then?si?numbers follow denoting the size of cakes in?i-th part. If there are multiple solutions, print any of them.

?

Sample Input

4 1 2 5 3 5 2 9 3

?

Sample Output

NO YES 1 5 2 1 4 2 2 3 NO YES 3 1 5 9 3 2 6 7 3 3 4 8

/**
hdu5355 思维+爆搜
题目大意：1~n这n个数分成m份。问能否成功，若能请给出一种分配方式
解题思路：假设sum(1~n)%m!=0无解，而且sum/m<n亦无解，其它情况有解。令ans=n%(2*m),若ans=0,则一个有效方案为(1+n),(2+n-1),(3+n-2)....
           每m个为一组，若ans！=0：令cnt=min(n,ans+(2*m)),这时候的cnt非常小我们爆搜就攻克了。关于爆搜我们总共要搜出m组。搜的时候枚举
           每组的最小的数，然后从该数開始往大了搜索
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
typedef long long LL;

vector<int>vec[55];
LL n,m,dis,ans;
int vis[105],pos[105],num[25];

bool dfs(int cnt,int sum,int id)///ans个数分成和相等的m份（id为第cnt份中最小的数。爆搜）
{
    if(cnt==m+1)return true;
    for(int i=ans;i>=id;i--)
    {
        if(vis[i])continue;
        if(sum+i==dis)
        {
            pos[i]=cnt;
            vis[i]=1;
            if(dfs(cnt+1,0,1))
                return true;
            vis[i]=0;
            return false;
        }
        else if(sum+i<dis)
        {
            pos[i]=cnt;
            vis[i]=1;
            if(dfs(cnt,sum+i,i+1))
                return true;
            vis[i]=0;
        }
    }
    return false;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld",&n,&m);
        LL sum=n*(n+1)/2;
        if(sum%m)
        {
            puts("NO");
            continue;
        }
        LL ave=sum/m;
        if(ave<n)
        {
            puts("NO");
            continue;
        }
        puts("YES");
        ans=n%(2*m);
        if(ans!=0)
        {
            ans+=2*m;
            ans=min(ans,n);
        }
        for(int i=1;i<=m;i++)
        {
            vec[i].clear();
        }
        memset(num,0,sizeof(num));
        for(int i=n;i>ans;i-=(2*m))///分成2*m份
        {
            for(int j=1;j<=m;j++)
            {
                int x=i-j+1;
                int y=i-2*m+j;
                vec[j].push_back(x);
                vec[j].push_back(y);
                num[j]+=(x+y);
            }
        }
        dis=ave-num[1];
        memset(vis,0,sizeof(vis));
        memset(pos,0,sizeof(pos));
        dfs(1,0,1);
        for(int i=1;i<=ans;i++)
        {
            vec[pos[i]].push_back(i);
        }
        for(int i=1;i<=m;i++)
        {
            printf("%d",vec[i].size());
            for(int j=0;j<vec[i].size();j++)
            {
                printf(" %d",vec[i][j]);
            }
            printf("\n");
        }
    }
    return 0;
}