Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24456 Accepted Submission(s): 16986
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!"
"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
Author
Ignatius.L
f[n][m]表示把整数n分为最大不超过整数为m的个数。显然f[n][1]=1,当n<m时,f[n][m]=f[n][n]
f[1][m]=1,当n=m时,f[n][m]=f[n][m-1]+1
当n>m时,可以按最大数是否为m分类,如果最大是m,有f[n-m][m],如果最大不是m,f[n][m-1].
不需要递归,直接二重循环遍历就行
#include<iostream> #define maxn 121 using namespace std; int dp[maxn][maxn]; int main() { int i,j,n; for(int i=1;i<=121;i++) dp[1][i]=dp[i][1]=1; for(int i=2;i<121;i++) { for(int j=2;j<=121;j++) { if(i<j) { dp[i][j]=dp[i][i]; } else if(i==j) { dp[i][j]=dp[i][j-1] + 1; } else { dp[i][j] = dp[i-j][j] + dp[i][j-1]; } } } while(cin>>n) { cout<<dp[n][n]<<endl; } }