hdu1028 poj1644放苹果poj3014 Cake Pieces and Plates----整数划分模板

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10679    Accepted Submission(s): 7569


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

经典的算法之一-------整数划分问题

可以用递归法，母函数法，

递归法

import java.util.Scanner;

public class hdu1028整数划分 {
static long[][]a=new long[122][122];

private static long fun(int n,int m) {
if(a
[m]!=0)return a
[m];
if(n<1||m<1)return 0;
if(n==1||m==1)return 1;
if(n<m) return a
[m]=fun(n,n);
if(n==m) return a
[m]=1+fun(n,n-1);
return a
[m]=fun(n,m-1)+fun(n-m,m);

}

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int n = sc.nextInt();
System.out.println(fun(n,n));
}
}
}

POJ1644放苹果

放苹果

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 24361		Accepted: 15493

Description

把M个同样的苹果放在N个同样的盘子里，允许有的盘子空着不放，问共有多少种不同的分法？（用K表示）5，1，1和1，5，1 是同一种分法。
Input

第一行是测试数据的数目t（0 <= t <= 20）。以下每行均包含二个整数M和N，以空格分开。1<=M，N<=10。
Output

对输入的每组数据M和N，用一行输出相应的K。
Sample Input

1
7 3

Sample Output

8

Source

lwx@POJ

import java.util.Scanner;

public class poj放苹果_整数划分变化 {
//1983210400	1664	Accepted	3000K	125MS	Java	581B	2013-11-27 14:45:43
/**
* @param args
*/
static int a,b;
static int[][]s=new int[25][25];
public static int f(int n,int m){
if(s
[m]!=0)return s
[m];
if(n<1||m<1) return 0;
if(n==1||m==1) return 1;
if(n<m) return s
[m]=f(n,n);
if(n==m)return s
[m]=1+f(n,n-1);
return s
[m]=f(n,m-1)+f(n-m,m);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
while(N-->0){
a=sc.nextInt();
b=sc.nextInt();
System.out.println(f(a,b));
}
}
}

DP方法：大牛提供

import java.util.Scanner;
public class poj1664放苹果_整数划分变化{
//网上看到的，动态规划方法
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int []dp;

int N=sc.nextInt();
while(N-->0){
int m=sc.nextInt();//苹果
int n=sc.nextInt();//盘子
dp=new int [50];
dp[0]=1;
for(int i=1;i<=n;i++)//盘子
for(int j=i;j<=m;j++){//苹果
dp[j]+=dp[j-i];

}
System.out.println(dp[m]);
}
}
}

POJ3014

Cake Pieces and Plates

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 6444		Accepted: 2013
Case Time Limit: 1000MS

Description

kcm1700, ntopia, suby, classic, tkwons, and their friends are having a birthday party for kcm1700. Of course, there is a very large birthday cake. They divide the birthday cake into undistinguishable pieces and put them on identical plates. kcm1700 is curious,
so he wants to know how many ways there are to put m cake pieces on n plates.

Input

In the only input line, there are two integers n, m (1 ≤ n, m ≤ 4 500), which are the number of the plates and the number of the cake pieces respectively.

Output

If the number of ways is K, just output K mod 1000000007 because K can be very large.

Sample Input

3 7

Sample Output

8

Hint

There are 8 ways to fill 3 plates with 7 cakes, namely (0,0,7), (0,1,6), (0,2,5), (0,3,4), (1,1,5), (1,2,4), (1,3,3), (2,2,3).

Source

POJ Monthly--2006.09.29, Kim, Chan Min (kcm1700@POJ)

import java.util.Scanner;
public class poj1664放苹果_整数划分变化{
//网上看到的，动态规划方法
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int []dp;

int N=sc.nextInt();
while(N-->0){
int m=sc.nextInt();//苹果
int n=sc.nextInt();//盘子
dp=new int [50];
dp[0]=1;
for(int i=1;i<=n;i++)//盘子
for(int j=i;j<=m;j++){//苹果
dp[j]+=dp[j-i];

}
System.out.println(dp[m]);
}
}
}