HDU 5355 Cake

Cake

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1160    Accepted Submission(s): 157
Special Judge

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

41 25 35 29 3

 

Sample Output

NOYES1 52 1 42 2 3NOYES3 1 5 93 2 6 73 3 4 8

 

Source

2015 Multi-University Training Contest 6

 

Recommend

wange2014

题意：给你n块蛋糕，重量分别是1，2，……n-1,n;分给m个人，每个人得到的重量相同；

解法：跟square那题一样，直接深搜，可能会爆栈，开头加上#pragma comment(linker, "/STACK:102400000,102400000") 就好了。

注意：n最大为10的五次，n^2会爆int，要用__Int64.

代码：

#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <string.h>int n, m;int p;int vis[111111];int ans[11][111111];int num[11];int now[111111];int dfs(int bian,int len,int s) {                                                  if(bian==m-1)           return 1;    for(int i=s;i<=n;i++)    {        if(!vis[i])        {            vis[i]=1;            if(len+now[i]<p)            {                if(dfs(bian,len+now[i],i+1))                  {                                ans[bian][num[bian]++] = now[i];                         return 1;                }            }            else if(len+now[i]==p)            {                if(dfs(bian+1,0,0))                  {                    ans[bian][num[bian]++] = now[i];                    return 1;                }            }            vis[i]=0;        }    }    return 0;}int main(){    int t;    scanf("%d", &t);    while(t--)    {                memset(vis,0,sizeof(vis));        memset(num,0,sizeof(num));        memset(ans,0,sizeof(ans));        scanf("%d%d", &n, &m);        for(int i = 0;i < n;i++)            now[i] = n - i;        __int64 sum = (__int64)(1 + n) * n / 2;        p = sum / m;        if(sum%m!=0)        {            printf("NO\n");            continue;        }        else        {            if(dfs(0,0,0))            {                printf("YES\n");                int temp = 0;                for(int i = 0;i< m - 1;i++)                {                    printf("%d",num[i]);                    temp += num[i];                    for(int j = 0;j<num[i];j++)                        printf(" %d",ans[i][j]);                    printf("\n");                    }                temp = n - temp;                printf("%d",temp);                for(int i = 0;i < n;i++)                    if(!vis[i])                        printf(" %d",now[i]);                printf("\n");                    }            else            {                printf("NO\n");            }        }    }}