hdu1028 Ignatius and the Princess III 递推

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17872    Accepted Submission(s): 12532

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

41020

 

Sample Output

542627

 

转载自：http://blog.sina.com.cn/s/blog_677a3eb30100kqnn.html

传送门

首先，我们引进一个小小概念来方便描述吧，record[n][m]是把自然数划划分成所有元素不大于m的分法，例如：
当n=4，m=1时，要求所有的元素都比m小，所以划分法只有1种：{1，1，1，1}；
当n=4，m=2时，。。。。。。。。。。。。。。。。只有3种{1，1，1，1},{2，1，1}，{2，2}；
当n=4，m=3时，。。。。。。。。。。。。。。。。只有4种{1，1，1，1},{2，1，1}，{2，2},{3,1}；
当n=4，m=5时，。。。。。。。。。。。。。。。。只有5种{1，1，1，1},{2，1，1}，{2，2},{3,1},{4}；
从上面我们可以发现：当n==1||m==1时，只有一种分法；
当n<m时，由于分法不可能出现负数，所以record[n][m]=record[n][n];
当n==m时，那么就得分析是否要分出m这一个数，如果要分那就只有一种{m}，要是不分，那就是把n分成不大于m-1的若干份；即record[n][n]=1+record[n][n-1];
当n>m时，那么就得分析是否要分出m这一个数，如果要分那就{{m},{x1,x2,x3..}}时n-m的分法record[n-m][m]，要是不分，那就是把n分成不大于m-1的若干份；即record[n][n]=record[n-m][m]+record[n][m-1];
那么其递归式：

dp[i][j]将整数i划分为最大不超过j的

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int main(){int N, dp[120 + 5][120 + 5];for (int i = 1; i <= 123; i++) dp[i][1] = dp[1][i] = 1;for (int i = 2; i <= 123; i++) {for (int j = 2; j <= 123; j++) {if (i == j) {dp[i][j] = 1 + dp[i][j - 1];}else if (i < j) {dp[i][j] = dp[i][i];}else if (i > j) {dp[i][j] = dp[i][j - 1] + dp[i - j][j];}}}while (~scanf("%d", &N)) printf("%d\n", dp[N][N]);return 0;}