HDU -> ACM -> Ignatius and the Princess III

Ignatius and the Princess III

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 40   Accepted Submission(s) : 30

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<stdio.h>  
#include<string.h>  
const int N = 125;  
#define LL __int64  
  
LL dp[N][N][N],sum[N][N];  
int main(){  
    memset(dp,0,sizeof(dp));  
    memset(sum,0,sizeof(sum));  
    for(int i=1;i<=124;i++)  
        dp[i][1][i]=1;  
    sum[1][1]=1;  
    for(int i=2;i<=124;i++)//苹果数  
    for(int j=2;j<=i;j++){  //盘子数  
        for(int k=0;k<=i/j;k++){ //最后一个盘放的数量  
            dp[i][j][k]=0;  
            for(int tk=k;tk<=(i-k)/(j-1);tk++)//倒数第二个盘放的数量  
            dp[i][j][k]+=dp[i-k][j-1][tk];  
            sum[i][j]+=dp[i][j][k];  
        }  
    }  
    int n;  
    while(scanf("%d",&n)>0)  
        printf("%I64d\n",sum[n][n]);  
}