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：

1 #include
      
        2 const int MAXN=10010; 3 int main(){ 4     int a[MAXN],b[MAXN],N; 5     while(~scanf("%d",&N)){ 6         int i,j,k; 7         for(i=0;i<=N;i++){ 8             a[i]=1;b[i]=0; 9         }10         for(i=2;i<=N;i++){11             for(j=0;j<=N;j++)12                 for(k=0;k+j<=N;k+=i)13                     b[j+k]+=a[j];14                 for(j=0;j<=N;j++)15                     a[j]=b[j],b[j]=0;16             }17         printf("%d\n",a[N]);18     }19     return 0;20 }

extern "C++"{#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            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;}