Cake ZQUOJ 25779

来源：互联网发布：税控盘开票软件下载编辑：程序博客网时间：2024/05/20 18:04

Cake

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 5 Accepted Submission(s): 2
Special Judge

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 s_i of i-th line is the number of cakes in i-th part. Then s_i 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

Source

2015 Multi-University Training 6

题意：给你n个尺寸大小分别为1,2,3,…,n的蛋糕，要求你分成m份，要求每份中所有蛋糕的大小之和均相同，如果有解，输出“YES”，并给出每份的蛋糕数及其尺寸大小，否则输出“NO”

例如n=5,m=3，即大小尺寸分别为1,2,3,4,5的5个蛋糕，要求分成三份，那么解可以是第一份一个蛋糕，大小为5；第二份两个蛋糕，大小为1、4；第三份两个蛋糕，大小为2、3。这样每份大小之和均为5，满足题目要求。

放上出题人的解题报告

解题思路：此题无解的情况无外乎两种：①所有蛋糕的大小之和无法被m整除；②每一份蛋糕的大小之和不得小于蛋糕大小的最大值n，即sum/m>=n,转化一下为n*m<=sum，亦可以将sum代入，转化为n>=2*m-1。

之后便是有解的情况，有解的情况每份蛋糕的大小之和是知道的，即sum/m。然后我们可以知道n最小得是2*m-1，举个例子就很清楚了，譬如n=5,m=3，

5 3

1 5

2 4 1

2 3 2

2*m-1就说明了n本身可以自己一组，其他两两一组，一个正向递增，一个反向递减，必定可以凑成n，当然这是奇数的情况

但该方法其实是不分奇偶性的，上述只是说明。其实我们需要做的就是把分配方式分成两部分，一部分是k个2*m，另外就是剩下的那部分，我们来看看n=23,m=6这组数据，之前AC的代码普遍过不了这组数据

23 6
YES
3 11   12 23
4 10 1  13 22
4   9 2  14 21
4   8 3  15 20
4   7 4  16 19
4   6 5  17 18

先利用(n-m*2+1)%(m*2)+m*2-1算出粉色那部分数据的大小，然后利用贪心的思想二分一下即可找到符合的蛋糕大小，剩下的绿色部分则是k个2*m的部分，该部分小的数是正向递增的，大的数是反向递减的，所以必定是符合要求的。为了更加直观，放上n=100，m=5的数据

100 5
YES
20 10 1  11 100  12 99  13 98  14 97  15 96 16 95 17 94 18 93 19 92
20 9 2  20 91  21 90  22 89  23 88  24 87 25 86 26 85 27 84 28 83
20 8 3  29 82  30 81  31 80  32 79  33 78 34 77 35 76 36 75 37 74
20 7 4  38 73  39 72  40 71  41 70  42 69 43 68 44 67 45 66 46 65

20 6 5 47 64 48 63 49 62 50 61 51 60 52 59 53 58 54 57 55 56

#include<bits/stdc++.h>using namespace std;int main(){    int t,r;    scanf("%d",&t);    while(t--)    {        long long n,m;        scanf("%lld %lld",&n,&m);        if((n*(n+1)/2)%m||2*m-1>n)            printf("NO\n");        else        {            printf("YES\n");            set<int> v;            set<int>::iterator it;            int w[32];            int c=(n-2*m+1)%(2*m)+2*m-1;            int s=(c*(c+1))/(2*m);            int d=(n-c)/(2*m);            for(int i=1;i<=c;i++)                v.insert(i);            for(int j=0,k=c+1;j<m;j++,putchar('\n'))            {                for(r=c=0;r<s;)                {                    it=v.upper_bound(s-r);                    r+=w[c++]=*--it;                    v.erase(it);                }                printf("%d",c+d*2);                for(int i=0;i<c;i++)                    printf(" %d",w[i]);                for(int i=0;i<d;i++)                    printf(" %d %d",k++,n--);            }        }    }}

0 0