HDOJ5355 Cake 构造

直接贪心的做法是不对的(例如: 23 6 , 27 7 , 28 7, 31 8 等...)

多校的时候因为SPJ的问题很多WA的程序都过了,实际上这题并没有那么水......

无解的有两种情况 1: sum不能被m整除 2: sum/m比n小

剩下的情况都有解:

标程的做法是根据: n(n+1)/(2*m) 每一组里都有k对组合,如第一行中的(11 100) (12 99) (13 98)....

因为有m行所以这样的组合有2*k*m个,这些对都可以凑成一样的大小

剩下的为粉红色的部分, 这些数字有(n-(2*m-1))%(2*m)+(2*m-1)个 (如样例  1~10),然后对这部分进行从大到小的贪心.....(感觉还是有问题???....)

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

参考题解: http://blog.csdn.net/queuelovestack/article/details/47321211

Cake

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1347    Accepted Submission(s): 197
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

   
   
   
   

    
    
    
    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 Contest 6

 

ac的代码:

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月07日 星期五 14时23分36秒
File Name     :HDOJ5355_2.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

LL n,m;
vector<int> vi[10];
set<int> v;

void init()
{
	for(int i=0;i<10;i++) vi[i].clear();
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);

	while(T_T--)
	{
		cin>>n>>m;
		init();
		LL sum=n*(n+1)/2;
		if(sum%m||n+1<2*m)
		{
			puts("NO"); continue;
		}

		LL c=(n-m*2+1)%(m*2)+m*2-1;  

		v.clear();
		for(int i=1;i<=c;i++)
		{
			v.insert(i);
		}

		LL s=c*(c+1)/(m*2);

		for(int i=0;i<m;i++)
		{
			set<int>::iterator it;
			LL ts=0;
			while(ts<s)
			{
				it=v.upper_bound(s-ts);
				LL tmp=*--it;
				vi[i].push_back(tmp);
				ts+=tmp;
				v.erase(it);
			}
		}

		int st=c+1;
		int ed=n;
		int k=(n-c)/(2*m);

		for(int i=0;i<m;i++)
		{
			for(int j=0;j<k;j++)
			{
				vi[i].push_back(st); st++;
				vi[i].push_back(ed); ed--;
			}
		}

		puts("YES");

		for(int i=0;i<m;i++)
		{
			int sz=vi[i].size();
			printf("%d",sz);
			for(int j=0;j<sz;j++)
			{
				printf(" %d",vi[i][j]);
			}
			putchar(10);
		}
	}
    
    return 0;
}

标程的代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int sol(){
  int n,m;
  scanf("%d%d",&n,&m);
  if(((n+1ll)*n>>1)%m!=0 || m * 2 -1 > n) return puts("NO");
  
  const int c=(n+1-m*2)%(m*2)+m*2-1,
    s=c*(c+1)/(m*2),
    d=(n-c)/(m*2);
  puts("YES");
  set<int> v;
  for(int i=1;i<=c;++i)v.insert(i);
  for(int j=0,l=c+1;j<m;++j,putchar('\n')){
    static int o[32],c,r;
    for(c=r=0;r<s;){
      auto it = v.upper_bound(s-r);
      r+=o[c++]=*--it;
      v.erase(it);
    }
    printf("%d",c+d*2);
    for(int i=0;i<c;++i)printf(" %d",o[i]);
    for(int i=0;i<d;++i)printf(" %d %d",l++,n--);
  }
}
int main(){
  int T;
  for(scanf("%d",&T);T--;sol());
}