HDOJ 1028 多种解法

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概述

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23401    Accepted Submission(s): 16310


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

题目大概意思：给定一个n，有几种分解方法。（4=1+3和4=3+1算一种）。

这道题最开始的思路是DP。但是后来了解到母函数，这又是一个陌生的知识点。哎，关于数学方面的知识还是太少了。无论是数论还是组合数学。不说废话。

第一种方法：
转换成背包问题。由于这道题数据比较小，N最大是120。所以可以理解成120容量的背包。1-120体积的物品任意把背包装满一共有多少种方法。
状态转移公式：dp[n] = dp[n]+dp[n-j]；

#include<bits/stdc++.h>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
using namespace std ;
typedef long long ll;
typedef unsigned long long ull;
const int MAX = 125;
int dp[MAX];
int main(){
    int n; 
    while(cin>>n){
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i = 1 ; i <= n ; i++)
            for(int j = i ; j <= n ; j++){
                dp[j]+=dp[j-i];
            }
        cout<<dp[n]<<endl;  
    }
    return 0;
}

第二种方法：
也是使用dp的方法。
dp[i][j] ：数i不超过j的分解方法数量。
dp[i][1] = 1;
i >= j
当i == j 时
dp[i][j]=dp[i][j-1]+1; （如果i==j那么其实就是比dp[i][j-1]多了一种j的分解方法）
当i > j 时
dp[i][j]=dp[i][j-1]+dp[i-j][t]; （dp[i][j]其实是在dp[i][j-1]的基础上多了j的划分。而j的划分，其实就是i-j一共有多少种分解方法。）；
上述t为min（i-j，j）因为这样才能使得dp有意义。

#include<bits/stdc++.h>
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
using namespace std ;
typedef long long ll;
typedef unsigned long long ull;
const int MAX = 125;
int dp[MAX][MAX];
int main(){
    int n; 
    while(cin>>n){
        memset(dp,0,sizeof(dp));
        for(int i = 1 ; i <= n ; i++)
            dp[i][1] = 1 ;
        for(int i = 2 ; i <= n ; i++)
            for(int j = 2 ; j <= i ; j++){
                if(i==j) dp[i][j] = dp[i][j-1]+1;
                else{
                    if (i-j >= j)
                    dp[i][j] = dp[i][j-1]+dp[i-j][j];
                    else 
                    dp[i][j] = dp[i][j-1]+dp[i-j][i-j];
                } 
            }
        cout<<dp[n][n]<<endl;   
    }
    return 0;
}