HDU 1028 Ignatius and the Princess III (母函数,或者DP)
2012-04-16 23:48
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Ignatius and the Princess III
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 2 Accepted Submission(s) : 2
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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
Author
Ignatius.L前面已经写了一篇用母函数做的:
/article/4679780.html
母函数的意思就是把n 用1,2,3,4,5`````来表示的种数,构造母函数:(1+x+x^2+x^3+`````)*(1+x^2+x^4+````)*(1+x^3+x^6+````)````
其实发现不用母函数也可以做出来。
就是用递归也可以做,只不过递归的时候要DP的方法记录下来。
程序如下:
#include<stdio.h> #include<string.h> const int MAXN=130; int dp[MAXN][MAXN]; //dp[i][j]表示 i 表示成最大的数不超过 j 的方法数 int calc(int n,int m) { if(dp [m]!=-1) return dp [m]; if(n<1||m<1) return dp [m]=0; if(n==1||m==1) return dp [m]=1; if(n<m) return dp [m]=calc(n,n); if(n==m) return dp [m]=calc(n,m-1)+1; return dp [m]=calc(n,m-1)+calc(n-m,m); } int main() { int n; memset(dp,-1,sizeof(dp)); while(scanf("%d",&n)!=EOF) printf("%d\n",calc(n,n)); return 0; }
这个的速度比母函数做要快。
而且想对好理解一点。
dp
[m]的意思是把n,用最大不超过m的数来表示的方法数。
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