hdu 1028Ignatius and the Princess III（dp）

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

41020

 

Sample Output

542627

 

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<iostream>#include<cstring>#include<string>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<<dp[n]<<endl;}return 0;} 

第二张方法可以用母函数   基本就是套模板了

#include<iostream>#include<cstring>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<<c1[nNUM]<<endl;}return 0;}