hdu---1028Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11046    Accepted Submission(s): 7821

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

 

 

/*    思路：    开设一个二维数组： f[n][m]代表数字n，分解为不超过m的多个数字，有多少种分法；采用递推的方式： for(i=1;i<130;i++)                    for(j=1;j<130;j++) 进行不同情况的递推；        对于数字i，分解为不超过j的多个数字的和。    每次递推，即看是否分出大小为m的数字；*/#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#define N 130int f[N][N];int main(){    int i,j,k=0;    freopen("in.txt","r",stdin);    f[1][1]=1;    for(i=1;i<N;i++)        for(j=1;j<N;j++)            if(i==j) f[i][j]=f[i][j-1]+1;            else if(i<j) f[i][j]=f[i][i];            else f[i][j]=f[i-j][j]+f[i][j-1];    while(scanf("%d",&k)!=EOF){        printf("%d\n",f[k][k]);    }    return 0;}