HDU Ignatius and the Princess III (母函數)
阿新 • • 發佈：2017-07-13
tex namespace class .net author def urn rst isp 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
Author Ignatius.L
1 /* 2 母函數模板 3 http://blog.csdn.net/vsooda/article/details/7975485代碼詳解 4 */ 5 #include<cstdio> 6 #include<iostream> 7 #define MAXN 121 8 9 using namespace std; 10 11 int c1[MAXN]; 12 int c2[MAXN]; 13 14 int n; 15 16 int main() { 17 while(~scanf("%d",&n)) { 18 for(int i=0;i<=n;i++) { 19 c1[i]=1;c2[i]=0; 20 } 21 for(int i=2;i<=n;i++) { 22 for(int j=0;j<=n;j++) 23 for(int k=0;k+j<=n;k+=i) 24 c2[j+k]+=c1[j]; 25 for(int j=0;j<=n;j++) { 26 c1[j]=c2[j]; 27 c2[j]=0; 28 } 29 } 30 printf("%d\n",c1[n]); 31 } 32 return 0; 33 }
HDU Ignatius and the Princess III (母函數)