[ACM] hdu 1028 Ignatius and the Princess III (母函数)
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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): 11475 Accepted Submission(s): 8118
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
41020
Sample Output
542627
Author
Ignatius.L
解题思路:
想了好半天,终于把母函数理解的差不多了。不同的题目需要有不同的母函数,该题的母函数为
(1+x+x^2+x^3...) * (1+x^2+x^4+x^6....) * (1+x^3+x^6+x^9.....)*(1+x^4+x^8+x^12.....)
1选择0,1,2,3个等 2选择0,1,2,3个等 3选择0,1,2,3个等
每个式子前面的1代表选择0个,即x^0
每个数都可以取无穷多个。
母函数的算法主要就是模拟手动计算多个式子相乘,记录的是每个指数的系数,比如 c[ 4 ]=2 指的是 x^4的系数为2
代码:
#include <iostream>using namespace std;int c[121],temp[121];int n;int main(){ while(cin>>n) { for(int i=0;i<=n;i++)//模拟多个式子相乘 { c[i]=1;//c[]中一开始是第一个式子各个指数的系数,即(1+x+x^2+x^3...),都为1 temp[i]=0; } for(int i=2;i<=n;i++)//表示和第几个式子相乘 { for(int j=0;j<=n;j++)//c[]中每一项的指数 for(int k=0;k+j<=n;k+=i)//第i个式子中的每一项的指数 temp[j+k]+=c[j];//临时指数 for(int j=0;j<=n;j++) { c[j]=temp[j];//每乘完一个式子,两个式子相当于合并成一个式子,并把系数存到c[]中 temp[j]=0; } } cout<<c[n]<<endl; } return 0;}
参考博客:http://www.wutianqi.com/?p=596
贴一下模板:
#include <iostream>using namespace std;// Author: Tanky Woo// www.wutianqi.comconst int _max = 10001;// c1是保存各项质量砝码可以组合的数目// c2是中间量,保存没一次的情况int c1[_max], c2[_max];int main(){//int n,i,j,k;int nNum; //int i, j, k;while(cin >> nNum){for(i=0; i<=nNum; ++i) // ---- ①{c1[i] = 1;c2[i] = 0;}for(i=2; i<=nNum; ++i) // ----- ②{for(j=0; j<=nNum; ++j) // ----- ③for(k=0; k+j<=nNum; k+=i) // ---- ④{c2[j+k] += c1[j];}for(j=0; j<=nNum; ++j) // ---- ⑤{c1[j] = c2[j];c2[j] = 0;}}cout << c1[nNum] << endl;}return 0;}
自己手动模拟了一遍才弄懂这个模板的意思。
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