Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22045 Accepted Submission(s): 15395
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
母函数
可以看博客： 链接 这里写链接内容
题目代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
int i,j,k,n,text[1010],text1[1010];
while(scanf("%d",&n)!=EOF)
{
memset(text1,0,sizeof(text1));
for(int i=0;i<=n;i++)
text[i]=1;
for(int i=2;i<=n;i++) //括号数
{
for(int j=0;j<=n;j++)//j和k是相乘时间双方的指数
{
for(int k=0;j+k<=n;k+=i)
text1[j+k]+=text[j];//数组储存的都是系数
}
for(int j=0;j<=n;j++)//交换 数组 将text1数组初始化
{
text[j]=text1[j];
text1[j]=0;
}
}
printf("%d
",text[n]);
}
return 0;
}大佬用 一维dp 写的：
/Hdu 1028
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
long long int a[100100];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<=n;i++)
a[i]=0;
a[0]=1;
for(int i=1;i<=n;i++)
{
for(int j=i;j<=n;j++)
{
a[j]+=a[j-i];
}
}
printf("%lld
",a[n]);
}
return 0;
}