POJ3014 Cake Pieces and Plates (整数划DP)

最新推荐文章于 2018-12-26 02:21:20 发布

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最新推荐文章于 2018-12-26 02:21:20 发布

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分类专栏： ACM 文章标签： output input

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本文介绍了一个经典的组合数学问题——如何将m块蛋糕分配到n个盘子中的算法实现。使用动态规划方法解决该问题，并提供了完整的C++代码示例。

Cake Pieces and Plates

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 5123		Accepted: 1424
Case Time Limit: 1000MS

Description

kcm1700, ntopia, suby, classic, tkwons, and their friends are having a birthday party for kcm1700. Of course, there is a very large birthday cake. They divide the birthday cake into undistinguishable pieces and put them on identical plates. kcm1700 is curious, so he wants to know how many ways there are to put m cake pieces on n plates.

Input

In the only input line, there are two integers n, m (1 ≤ n, m ≤ 4 500), which are the number of the plates and the number of the cake pieces respectively.

Output

If the number of ways is K, just output K mod 1000000007 because K can be very large.

Sample Input

3 7

Sample Output

Hint

There are 8 ways to fill 3 plates with 7 cakes, namely (0,0,7), (0,1,6), (0,2,5), (0,3,4), (1,1,5), (1,2,4), (1,3,3), (2,2,3).

题意很简单，就是求将m划分成为n个数的和，问有多少种方法，还是DP吧，呵呵。。。。。。。。

动态方程是：F(i,j)=F(i-j,j)+(i,j-1);

贴上代码：

#include <iostream> using namespace std; int n,m,dp[4501][4501]; int solve() { int i,j; for(i=1;i<=n;i++) dp[0][i]=1; for(i=1;i<=m;i++) { dp[i][0]=0; for(j=1;j<=n;j++) if(j>i) dp[i][j]=dp[i][j-1]; else dp[i][j]=(dp[i-j][j]+dp[i][j-1])%1000000007; } return dp[m][n]; } int main() { while(scanf("%d%d",&n,&m)!=EOF) printf("%d/n",solve()); return 0; }