Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4800 Accepted Submission(s): 3362
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
这个题有人说是DP，但是我最近在看母函数，所以用母函数过掉了
这个应该是母函数的入门题目了吧。。
我的代码：
#include<stdio.h>
int c1[125];
int c2[125];
int main()
{
int n,i,j,k;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<=n;i++)
{
c1[i]=1;
c2[i]=0;
}
for(i=2;i<=n;i++)
{
for(j=0;j<=n;j++)
for(k=0;k+j<=n;k=k+i)
c2[k+j]=c2[k+j]+c1[j];
for(j=0;j<=n;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
printf("%d\n",c1[n]);
}
return 0;
}