题目:HDU1028
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26149 Accepted Submission(s): 18059
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
方法:
此题应用母函数来解决,母函数就是常常用来解决组合方面的题目。母函数相关内容讲解:讲解1 讲解2
代码:
#include <iostream>
using namespace std;
int main()
{
int result[121];
int temp[121];
int target;
while (cin >> target)
{
for (int i = 0; i <= target; i++)
{
result[i] = 1;
temp[i] = 0;
}
for (int k = 2; k <= target; k++)
{
for (int i = 0; i <= target; i++)
for (int j = 0; j + i <= target; j += k)
temp[i+j] += result[i];
for (int i = 0; i <= target; i++)
{
result[i] = temp[i];
temp[i] = 0;
}
}
cout << result[target] << endl;
}
return 0;
}