hdu1028(dp 或母函数)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24446    Accepted Submission(s): 16980

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

Author

Ignatius.L

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题意：给你一个数n,问n有多少总凑法让这些数的和等于n

例如 3=1+1+1=1+2=3 有三种

题解：

1.母函数模板（第二类问题）

2.dp

显然凑数的时候大的数方法会覆盖小的数，那么凑大的数就可以往前找

例如凑5每次加一的话就往4加，加二就往3加

#include <iostream> using namespace std; int main (void) { int i,j,d[125]={1};//d[0]=1，其他自动初始化为0  for(i=1;i<121;i++) //从小到大加不会重复，从1往后加     for(j=i;j<121;j++) //从能+i的最小的那个数开始+n        d[j]+=d[j-i];//加到自身就是加d[0] （ =1 ）  while(cin>>i) cout<<d[i]<<endl; return 0; }