hdu 1028 Ignatius and the Princess III

March 13, 2013
Hdu DP

Ignatius and the Princess III

Time Limit: ²⁰⁰⁰⁄₁₀₀₀ MS (Java/Others) Memory Limit: ⁶⁵⁵³⁶⁄₃₂₇₆₈ K (Java/Others) Total Submission(s): 8447 Accepted Submission(s): 6009

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

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main(void){
#ifndef ONLINE_JUDGE
  freopen("1028.in", "r", stdin);
#endif
  int dp[121][121];
  memset(dp, 0, sizeof(dp)); 
  for (int i = 0; i < 121; ++i){
    dp[i][1] = dp[1][i] = 1;
  }
  for (int i = 2; i < 121; ++i){
    for (int j = 2; j < 121; ++j){
      if (i < j) dp[i][j] = dp[i][i];
      else if (i == j) dp[i][j] = 1 + dp[i][j-1];
      else dp[i][j] = dp[i][j-1] + dp[i-j][j];
    }
  }
  int n; 
  while (~scanf("%d", &n)){
    printf("%d\n", dp[n][n]);
  }
  return 0;
}

突然感觉动态规划特别有意思……

这道题目也可以用母函数做，可惜看了一下，没有看懂……

dp[i][j] 表示把整数i用不大于j的正整数之和表示.

分三种情况：

当ｉ》ｊ的时候，如果选ｊ，那么有dp[i-j][j]种；如果不选ｊ，那么有ｄｐ[ｉ][ｊ-1]种；

当ｉ==ｊ的时候，如果选ｊ，那么有1种；如果不选ｊ，那么又ｄｐ[ｉ][ｊ-1]种；

当ｉ《ｊ的时候，则ｄｐ[i][j] = dp[i][i]；

然后，显然，当ｉ或者ｊ其中一个等于1的时候，结果都是1.

求ｎ的划分，就是求把整数ｎ用不大于ｎ的正整数之和表示，有几种方法，就是ｄｐ[n][n]。

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