Ignatius and the Princess III （HDU 1028） ——母函数(另解DP)

        Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11999    Accepted Submission(s): 8494

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

41020

Sample Output

542627

这是一个典型的母函数问题（当然还有其他解法，比如说背包）

下面是这题的解题代码，也是母函数的模板型代码：

#include<stdio.h>int main(){    int n,i,j,k;    int c1[1000],c2[1000];    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[k+j]+=c1[j];            for(j=0; j<=n; j++)                c1[j]=c2[j],c2[j]=0;        }        printf("%d\n",c1[n]);    }    return 0;}

下面是别人对代码的一些解释：

#include<iostream>#include<stdio.h>using namespace std;int main(){    int N;    int c1[125],c2[125];    while(cin>>N)    {        int i,j,k;        for(i=0;i<=N;i++)//初始化第一个表达式的系数        {            c1[i]=1;            c2[i]=0;        }        for(i=2;i<=N;i++)        {//从第二个表达式开始，因为有无限制个，所以有n个表达式            for(j=0;j<=N;j++)            {//从累乘的表达式后的一个表达式第一个到最后一个                for(k=0;k+j<=N;k+=i)                {//k为第j个变量的指数，第i个表达式每次累加i                    c2[j+k]+=c1[j];                }            }            for(j=0;j<=N;j++)            {//滚动数组算完一个表达式后更新一次                c1[j]=c2[j];                c2[j]=0;            }        }        printf("%d\n",c1[N]);    }    return 0;}

母函数是一类解决数学中组合问题的方法，它有很多种，不懂可以百度一下“母函数”

另解:
 这是有关计数问题的DP中的划分数. 相当于n个无区别的物品,将它们划分成不超过m组(当然这里m=n),求划分方法数. 考虑n的m划分a1+a2+...+am = n, 如果对于每个i都有ai>0, 那么{ai-1}就对应了n-m的m划分. 另外,如果存在ai=0, 那么这就对应啦n的m-1划分. dp[i][j] : j的i划分的总数. 综上,我们可得出如下递推关系: dp[i][j] = dp[i][j-i] + dp[i-1][j].

/*************************************************************************    > File Name: dp.cpp    > Author: tj    > Mail: [e-mail]    > Created Time: 2015年01月02日 星期五 14时24分28秒 ************************************************************************/#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int main(){int n, dp[130][130];while(~scanf("%d%d", &n,&m)){memset(dp, 0, sizeof(dp));dp[0][0] = 1;for(int i=1; i<=n; i++){for(int j=0; j<=m; j++){if(j >= i) dp[i][j] = dp[i][j-i]+dp[i-1][j];else dp[i][j] = dp[i-1][j];}}printf("%d\n", dp[m][n]);}return 0;}