hdu 1028

                                                Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18994 Accepted Submission(s): 13352


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

这道题上算法课时学过，就是求一个整数n的划分种类，这里需要注意一点，我们不能使用递归式，会超时。只能用动态规划来进行打表。

<span style="font-size:18px;">#include <iostream>#include<cstdio>using namespace std;int dp[125][125];void init(){    for (int i = 1;i <= 120;i++)        dp[0][i] = 1;    for (int i = 1;i <= 120;i++)    {        dp[i][0] = 0;        for (int j = 1;j <= 120;j++)        {            if(i < j)                dp[i][j] = dp[i][i];            else                dp[i][j] = (dp[i-j][j]+dp[i][j-1]);        }    }}int main(){    int n;    init();    while(~scanf("%d",&n))    {        printf("%d\n",dp[n][n]);    }    return 0;}</span>

另外给上递归算法，非递归的就照着写就可以了

#include <iostream>#include<cstdio>using namespace std;int q(int n,int m){    if((n<1)||(m<1))        return 0;    if((n==1)||(m==1))        return 1;    if(n<=m)        return(q(n,n-1)+1);    return q(n,m-1)+q(n-m,m);}int main(){    int n;    scanf("%d",&n);    printf("%d\n",q(n,n));    return 0;}