HDU1028 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): 25867    Accepted Submission(s): 17879

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

  

G(x) = (1 + x^1 + x^2 + x^3 ......) * (1 + x^2 + x^4 + x^8.....) * (1 + x^3 + x^6 + x^9 ......)...........

x^k的系数即为数字k的拆分方案数

 

#include <bits/stdc++.h>
using namespace std;
int c1[125],c2[125];
int n;

int build(){
    for (int i = 0;i <= 120;++i){
        c1[i] = 1,c2[i] = 0;
    }
    for (int i = 2;i <= 120;++i){
        if (c1[i] == 0) return 0 * printf("%d fuck",i);
        for (int j = 0;j <= 120;++j){
            for (int k = 0;k+j <= 120;k+=i){
                    c2[k+j] += c1[j];
            }
        }
        for (int k = 0;k <= 120;++k){
                c1[k] = c2[k];
                c2[k] = 0;
        }
    }
}


int main(){
    build();
    while(~scanf("%d",&n)){
        printf("%d
",c1[n]);
    }
}

同类题目：HDU1398

#include <bits/stdc++.h>
using namespace std;
int c1[305],c2[305];
int n;

void calc(){
    for (int i = 0;i <= 302;++i) c1[i] = 1,c2[i] = 0;
    for (int k = 2;k <= 17;++k){
        for (int i = 0;i <= 302;++i){
            for (int j = 0;j + i <= 302;j += k*k){
                c2[j+i] += c1[i];
            }
        }
        for (int i = 0;i <= 302;++i) c1[i] = c2[i],c2[i] = 0;
    }
}

int main(){
    calc();
    while(~scanf("%d",&n) && n ){
        printf("%d
",c1[n]);
    }
}