HDU1028——Ignatius and the Princess III（背包问题dp）

Ignatius and the Princess III

活水源头

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18990    Accepted Submission(s): 13348

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
解题思路：
     具体看各种背包问题：各种背包等你来拿
代码实现:
#include <iostream>#include <cstdio>#include <cmath>#include <queue>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 1000000+10;int n;int dp[150][150];int main(){    while( ~scanf("%d",&n) )    {        memset(dp,0,sizeof(dp));        dp[1][1] = 1;        for( int j=0; j<=n; j++ )        {            dp[0][j] = dp[j][0] = 0;        }        for( int i=1; i<=n; i++ )        {            for( int j=1; j<=n; j++ )            {                if(i==j) dp[i][j] = dp[i][j-1] +1;                else if(i>j) dp[i][j] = dp[i-j][j] + dp[i][j-1];                else if(i<j) dp[i][j] = dp[i][i];            }        }        //for( int i=0; i<=n; i++ )        //{        //    for( int j=0; j<=n; j++ )        //        cout << dp[i][j] << " ";        //    cout <<endl;        //}        printf("%d\n",dp[n][n]);    }    return 0;}