HDU 1028 Ignatius and the Princess III（DP，整数划分）

最新推荐文章于 2021-02-19 10:33:34 发布

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最新推荐文章于 2021-02-19 10:33:34 发布

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分类专栏： DP

本文链接：https://blog.youkuaiyun.com/charis_lee/article/details/77278643

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本文介绍了一个经典的整数划分问题，通过动态规划方法求解给定正整数N的不同划分方式的数量。文章提供了完整的代码实现，并解释了状态转移方程。

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Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22498 Accepted Submission(s): 15713

Problem Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
  N=a[1]+a[2]+a[3]+...+a[m];
  a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
  4 = 4;
  4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

Sample Input

4
10
20

Sample Output

5
42
627

Author

Ignatius.L

刚开始的时候，我一直以为这是一道纯数学规律题，天真的以为可以一行公式代码ac，后来发现确实找不到规律QAQ；

再仔细分析一下题目，觉得这应该是一道DP题，因为大数的结果是和其往下分的小数密切相关的，所以就成了一道整数划分题；

状态转移方程：dp[i][j]=dp[i][j-1]+dp[i-j][j];(dp[i][j]表示将i进行划分，最大的数不超过j的方案个数）；

于是题目变得非常简单。

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
int dp[122][122];
int main(){
    for(int i=0;i<122;i++)
        dp[i][1]=dp[1][i]=dp[0][i]=1;
    for(int i=2;i<122;i++){
        for(int j=2;j<122;j++){
            if(j<=i)
                dp[i][j]=dp[i][j-1]+dp[i-j][j];
            else
                dp[i][j]=dp[i][i];
        }
    }
    int n;
    while(~scanf("%d",&n)){
        printf("%d\n",dp[n][n]);
    }
    return 0;
}