Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15659    Accepted Submission(s): 11038

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

 

Author

Ignatius.L

完全背包问题。把总和的那个数n看作是背包总容量。材料有1,2……n，这些材料向背包填充。最大价值对应的是最多方案。

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;int main(){   int n,i,j,a[150],dp[150];   while(scanf("%d",&n)!=EOF)   {       memset(dp,0,sizeof(dp));       dp[0]=1;      for(i=1;i<=n;i++)          a[i]=i;//材料容量      for(i=1;i<=n;i++)      {          for(j=a[i];j<=n;j++)          {              dp[j]+=dp[j-a[i]];          }      }      cout<<dp[n]<<endl;   }    return 0;}