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[计数dp] ural 1114. Boxes

2024-07-03 20:27:02   227人阅读

题目链接：

http://acm.timus.ru/problem.aspx?space=1&num=1114

1114. Boxes

Time limit: 0.6 second
Memory limit: 64 MB
N boxes are lined up in a sequence (1 ≤ N ≤ 20). You have A red balls and B blue balls (0 ≤ A ≤ 15, 0 ≤ B ≤ 15). The red balls (and the blue ones) are exactly the same. You can place the balls in the boxes. It is allowed to put in a box, balls of the two kinds, or only from one kind. You can also leave some of the boxes empty. It‘s not necessary to place all the balls in the boxes. Write a program, which finds the number of different ways to place the balls in the boxes in the described way.

Input

Input contains one line with three integers N, A and B separated by space.

Output

The result of your program must be an integer written on the only line of output.

Sample

inputoutput
2 1 1
9
Problem Source: First competition for selecting the Bulgarian IOI team.
Tags: none  (
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Difficulty: 194    Printable version    Submit solution    Discussion (29)
My submissions    All submissions (7145)    All accepted submissions (2115)    Solutions rating (1803)

题目意思：

有n个盒子，有相同的A球a个，相同的B球b个，每个盒子可以不放球可以放多个球，球可以不全放完。求放的种数。

n<=20 a,b<=15

解题思路：

简单计数dp

dp[i][j][k]:表示1~i个盒子中放了j个A球,k个B球的总的种数。

显然dp[i][j][k]=∑(dp[i-1][jj][kk]) (jj<=j,kk<=k)

注意要用usigned long long不然最后一个会超

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define ull unsigned long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 22


ull dp[Maxn][Maxn][Maxn];
int n,a,b;

void init()
{
    memset(dp,0,sizeof(dp));
    dp[0][0][0]=1;

    for(int i=1;i<=20;i++)
    {
        for(int j=0;j<=15;j++)
        {
            for(int k=0;k<=15;k++)
            {

                for(int jj=0;jj<=j;jj++)
                    for(int kk=0;kk<=k;kk++)
                        dp[i][j][k]+=dp[i-1][jj][kk];

            }
        }
    }

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

   while(~scanf("%d%d%d",&n,&a,&b))
   {
       //printf("%I64d\n",dp[n][a][b]);
       ll ans=0;

       for(int i=0;i<=a;i++)
            for(int j=0;j<=b;j++)
                ans+=dp[n][i][j];
       printf("%I64u\n",ans);

   }

   return 0;
}



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