Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17804 Accepted Submission(s): 12482

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

Sample Output

Author

Ignatius.L

i的变化有n种可能性 dp数组把各种可能性累加 j是i的变化情况

#include
    
     #include
     
      #include
      
       using namespace std;int dp[1000];int main(){		int i,j,k,u,n,m;	while(cin>>n)	{		memset(dp,0,sizeof(dp));		dp[0]=1;	for(i=1;i<=n;i++)		{		for(j=i;j<=150;j++)		{			dp[j]+=dp[j-i];		}	}		cout<
       
        <

第二张方法可以用母函数基本就是套模板了

#include
    
     #include
     
      using namespace std;const int _max=10001;int c1[_max],c2[_max];int main() {	int nNUM;	int i,j,k;	while(cin>>nNUM)	{	for(i=0;i<=nNUM;i++)	{		c1[i]=1;		c2[i]=0;	}		for(i=2;i<=nNUM;i++)	{		for(j=0;j<=nNUM;j++)		{			for(k=0;k+j<=nNUM;k+=i)			{				c2[k+j]+=c1[j];			}		}		for(j=0;j<=nNUM;j++)		{		c1[j]=c2[j];		c2[j]=0;			}	}	cout<
      
       <