hdu 1028 Ignatius and the Princess III(母函数)
2015-04-18 21:46
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Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14796 Accepted Submission(s): 10421
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
题目分析:
母函数模板提
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #define MAX 157 using namespace std; typedef long long LL; LL a[MAX],b[MAX]; int n; int main ( ) { memset ( a , 0 , sizeof ( a ) ); for ( int i = 0 ; i <= 120 ; i++ ) a[i] = 1; for ( int i = 2 ; i <= 120 ; i++ ) { memset ( b , 0 , sizeof ( b ) ); for ( int j = 0 ; j <= 120 ; j++ ) for ( int k = 0 ; k*i <= 120 ; k++ ) if ( k*i+j <= 120 ) b[k*i+j] += a[j]; for ( int j = 0 ; j <= 120 ; j++ ) a[j] = b[j]; } while ( ~scanf ( "%d" , &n ) ) { printf ( "%lld\n" , a ); } }
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