hdu5355 思维+爆搜

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，问是否能成功，若能请给出一种分配方式
解题思路：如果sum(1~n)%m!=0无解，并且sum/m
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;

vectorvec[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+ians;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