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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 思維+爆搜
題目大意：1~n這n個數分成m份。問能否成功，若能請給出一種分配方式
解題思路：假設sum(1~n)%m!=0無解，而且sum/m<n亦無解，其它情況有解。令ans=n%(2*m),若ans=0,則一個有效方案為(1+n),(2+n-1),(3+n-2)....
           每m個為一組，若ans！=0：令cnt=min(n,ans+(2*m)),這時候的cnt非常小我們爆搜就攻克了。關於爆搜我們總共要搜出m組。搜的時候枚舉
           每組的最小的數，然後從該數開始往大了搜索
*/
#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個數分成和相等的m份（id為第cnt份中最小的數。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;
}

hdu5355 思維+爆搜