HDU1028

HDU1028
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21783 Accepted Submission(s): 15208

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

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**题目分析：
拆分整数，拆为为n个整数的和的方案有多少种，重复无效。一开始知道推递推式用动态规划做，后面了解到了母函数；一个求多重组合类型题的函数。
**

//母函数法#include <iostream>using namespace std;const int N = 121;int c1[N], c2[N];int main(){    int n, i, j, k;    while (cin >> n)    {        memset(c1, 0, sizeof(c1));        memset(c2, 0, sizeof(c2));        for (i = 0; i <= n; i++)//i的作用是表示所求的x^i            c1[i] = 1;//c1[i]即x^i的系数，也是对用n=i的输出结果，这里求出了当i=1的值        for (i = 2; i <= n; i++)        {            for (j = 0; j <= n; j++)//j的作用是当前括号里的x^j                for (k = 0; k + j <= n; k += i)//k的作用是表示当前可用于计算的x^k                                            //例如当n=4，i=3时4=3+1                                            //k+j>n求出无意义，所以限制                {                    c2[j + k] += c1[j];//只有j有系数c1[j]                }            for (j = 0; j <= n; j++)            {                c1[j] = c2[j];                c2[j] = 0;            }        }        cout << c1[n] << endl;    }    return 0;}

拆分整数就是把整数n拆分成不大于m的数之和即cf[n][m]
由题目举例：
当n=3，m=1时，只有1种：{1，1，1}；cf[3][1]=1;
当n=3，m=2时，只有2种{1，1，1},{2，1}；cf[3][2]=2;
当n=3，m=3时，只有3种{1，1，1},{2，1}，{3}；cf[3][3]=3;
当n=3，m=4时，只有3种{1，1，1},{2，1}，{3}；cf[3][4]=3;
当n=4，m=1时，只有1种：{1，1，1，1}；cf[4][1]=1;
当n=4，m=2时，只有3种{1，1，1，1},{2，1，1}，{2，2}；cf[4][2]=3;
当n=4，m=3时，只有4种{1，1，1，1},{2，1，1}，{2，2},{3,1}；cf[4][3]=4;
当n=4，m=4时，只有5种{1，1，1，1},{2，1，1}，{2，2},{3,1},{4}；cf[4][4]=5
当n=4，m=5时，只有5种{1，1，1，1},{2，1，1}，{2，2},{3,1},{4}；cf[4][5]=5
从上面我们可以发现：
当n==1||m==1时，只有一种分法；
当n< m时,cf[n][m]=cf[n][n];
当n==m时，cf[n][n]=1+cf[n][n-1];
当n>m时，cf[n][m]=cf[n-m][m]+cf[n][m-1];

//动态规划#include<iostream>using namespace std;int cf[122][122] = { 0 };int main(){    int i, j;    for (i = 1; i <= 121; i++)        cf[1][i] = cf[i][1] = 1;    for (i = 2; i<121; i++)    {        for (j = 2; j <= 121; j++)        {            if (i<j)                cf[i][j] = cf[i][i];            else if (i == j)                cf[i][j] = 1 + cf[i][j - 1];            else if (i>j)                cf[i][j] = cf[i - j][j] + cf[i][j - 1];        }    }    int n;    while (cin >> n)        cout << cf[n][n] << endl;    return 0;}