HDU-1028-Ignatius and the Princess III（完全背包）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1028

Ignatius and the Princess III

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

题目分析：整数划分问题是将一个正整数n拆成一组数连加并等于n的形式，且这组数中的最大加数不大于n。和上一题的钱币兑换问题类似。完全背包。

代码如下：

#include<iostream>#include<cstring>using namespace std;int main(){int n,dp[121];while(cin>>n){memset(dp,0,sizeof(dp));dp[0]=1;//递推 for(int i=1;i<=n;i++){for(int j=i;j<=n;j++){dp[j]=dp[j]+dp[j-i];}}cout<<dp[n]<<endl;} }