hdu 1028 Ignatius and the Princess III 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): 24975    Accepted Submission(s): 17253

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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问整数n有多少种拆分可能，那也就是求x^n的系数

所以我们直接用母函数求x^n的系数就行

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iterator>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 2*1e2 + 10;
const int mod = 10000;
typedef long long ll;
int main() {
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    ll n;
    while( cin >> n ) {
        ll a[maxn], b[maxn];
        for( ll i = 0; i <= n; i ++ ) {
            a[i] = 1, b[i] = 0;
        }
        for( ll i = 2; i <= n; i ++ ) { //最低是从2开始划分
            for( ll j = 0; j <= n; j ++ ) {
                for( ll k = 0; k*i+j <= n; k ++ ) {
                    b[k*i+j] += a[j];
                }
            }
            for( ll j = 0; j <= n; j ++ ) {
                a[j] = b[j], b[j] = 0;
            }
        }
        cout << a[n] << endl;
    }
    return 0;
}