Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15942 Accepted Submission(s): 11245
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.
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
10
20
Sample Output
5
42
627
42
627
Author
Ignatius.L
Recommend
一开始自己想了一种解法，类似dp，但是应该不是dp，应该算找规律，速度没dp快，因为多了一层循环，虽然最里面一层循环很小，
#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 130
int n,d[N][N];//d[i][j]表示组成不超过j的数组成i有多少种方法 int main()
{
for(int i=;i<=;i++)d[i][]=;
d[][]=;
for(int i=;i<=;i++)
{
for(int j=;j<=i;j++)
{
for(int k=j;k>=;k--)
{
d[i][j]+=d[i-k][min(i-k,k)];
}
}
}
while(~scanf("%d",&n))
{
cout<<d[n][n]<<endl;
}
return ;
}
看了网上的正规dp解法，稍加改进
#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 130
int n,d[N][N]; int main()
{
for(int i=;i<=;i++)d[i][]=;
d[][]=;
for(int i=;i<=;i++)
{
for(int j=;j<=i;j++)
{
d[i][j]=d[i][j-]+d[i-j][min(j,i-j)];
}
}
while(~scanf("%d",&n))
{
cout<<d[n][n]<<endl;
}
return ;
}
还有一种母函数的做法
以后再学习