HDU 1028 Ignatius and the Princess III(母函数)
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9408 Accepted Submission(s): 6642
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
题目分析:题目要求的是整数的拆分方案个数,用母函数处理。
母函数详解链接:http://www.wutianqi.com/?p=596
1 #include <stdio.h> 2 3 int N; 4 const int MAX_NUM = 125; 5 int solNum[MAX_NUM]; 6 int temp[MAX_NUM]; 7 8 void InitGenerFun() 9 { 10 for(int i = 0; i < MAX_NUM; i++) 11 { 12 solNum[i] = 1; 13 temp[i] = 0; 14 } 15 for(int i = 2; i < MAX_NUM; i++) 16 { 17 for(int j = 0; j < MAX_NUM; j++) 18 { 19 for(int k = 0; k+j < MAX_NUM; k+=i) 20 { 21 temp[k+j] += solNum[j]; 22 } 23 } 24 for(int k = 0; k < MAX_NUM; k++) 25 { 26 solNum[k] = temp[k]; 27 temp[k] = 0; 28 } 29 } 30 } 31 32 int main() 33 { 34 InitGenerFun(); 35 36 while(scanf("%d", &N) != EOF) 37 { 38 printf("%d\n", solNum[N]); 39 } 40 41 return 0; 42 }