970
HDU1028-Ignatius and the Princess III
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12789 Accepted Submission(s): 9045
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
赤裸裸的母函数
（有没有发现，大部分母函数的题基本都是模板题= =）
AC代码：
#include<stdio.h>
#include<string.h>
#define MAX 200
int c1[MAX],c2[MAX];
int main()
{
int i,j,n,k;
while(~scanf("%d",&n))
{
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+=i)
{
c2[j+k]+=c1[j];
}
for(j=0;j<=n;j++)
{
c1[j]=c2[j];
c2[j]=0;
}
}
printf("%d\n",c1[n]);
}
return 0;
}
最后更新：2017-04-03 05:39:49