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): 16028    Accepted Submission(s): 11302

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

 题解：相当于取砝码，每个砝码可以取多次；

4=1+1+1+1=1+1+2=1+3=2+2；

则母函数为：

相当于1的砝码取法从0....n,2的砝码0....n/2。。。。。。k的砝码从0....n/k；

代码：

#include<stdio.h>
const int MAXN=10010;
int main(){
    int a[MAXN],b[MAXN],N;
    while(~scanf("%d",&N)){
        int i,j,k;
        for(i=0;i<=N;i++){
            a[i]=1;b[i]=0;
        }
        for(i=2;i<=N;i++){
            for(j=0;j<=N;j++)
                for(k=0;k+j<=N;k+=i)
                    b[j+k]+=a[j];
                for(j=0;j<=N;j++)
                    a[j]=b[j],b[j]=0;
            }
        printf("%d\n",a[N]);
    }
    return 0;
}

 

extern "C++"{
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
const int INF = 0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
typedef unsigned long long ULL;

void SI(int &x){scanf("%d",&x);}
void SI(double &x){scanf("%lf",&x);}
void SI(LL &x){scanf("%lld",&x);}
void SI(char *x){scanf("%s",x);}

}
const int MAXN = 200;
int a[MAXN],b[MAXN];
int main(){
    int N;
    while(~scanf("%d",&N)){
        for(int i = 0;i <= N;i++)a[i] = 1,b[i] = 0;
        for(int i = 2;i <= N;i++){
            for(int j = 0;j <= N;j++){
                for(int k = 0;j + k <= N;k += i){
                    b[j + k] += a[j];
                }
            }
            for(int j = 0;j <= N;j++)a[j] = b[j],b[j] = 0;
        }
        printf("%d\n",a[N]);
    }
    return 0;
}