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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int main(void){
 6 #ifndef ONLINE_JUDGE
 7   freopen("1028.in", "r", stdin);
 8 #endif
 9   int dp[121][121];
10   memset(dp, 0, sizeof(dp)); 
11   for (int i = 0; i < 121; ++i){
12     dp[i][1] = dp[1][i] = 1;
13   }
14   for (int i = 2; i < 121; ++i){
15     for (int j = 2; j < 121; ++j){
16       if (i < j) dp[i][j] = dp[i][i];
17       else if (i == j) dp[i][j] = 1 + dp[i][j-1];
18       else dp[i][j] = dp[i][j-1] + dp[i-j][j];
19     }
20   }
21   int n; 
22   while (~scanf("%d", &n)){
23     printf("%d\n", dp[n][n]);
24   }
25   return 0;
26 }

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]。

posted on 2013-03-13 21:09 aries__liu 阅读(131) 评论(0) 编辑收藏举报

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Ignatius and the Princess III