HDU 1028 Ignatius and the Princess III(母函数)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15794    Accepted Submission(s): 11138

Description

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"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!"

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Input

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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.

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Output

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For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

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Sample Input

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4

10

20

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Sample Output

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5

42

627

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Analysis

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　　这是一道基础的母函数题

　　这一题就是算出 ( x^0 + x^1 + x^2 + ...)( x^0 + x^2 + x^4 + ...)( x^0 + x^3 + x^6 + ...) ...

　　如第一个括号中的 x^0 代表取0个1， x^2 代表取2个1, 价值 2*1 = 1

　　　第二个括号中的 x^0 代表取0个2， x^4 代表取2个2, 价值 2*2 = 4

　　指数代表取了之后的价值和

　　最后的展开式中某一项 a*x^b 代表 价值（总和）为 b 时有 a 种取法

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#include <cstdio>
#include <cstring>

int c1[125],c2[125];                    // c1[i] 表示指数为i的x的系数 即 a*x^3 中的a
                                        // c2[i] 表示两个因子相乘指数和为i时，乘积的系数 如 3*x^4 * x^5 = 3 * x^9 后面的3就是c2[9]

int main()
{
    int n;
    int i,j,k;

    while(~scanf("%d",&n))
    {
        for(i=0;i<=n;i++)              // 可以看做计算第一个括号 ( x^0 + x^1 + x^2 + ...) 每一项系数为1
        {
            c1[i] = 1;
            c2[i] = 0;
        }

        for(i=2;i<=n;i++)              // 计算 （前面已计算过的括号所累乘得到的 大大的括号）*（后面一个括号）
        {                              // 第一次循环就是计算 ( x^0 + x^1 + x^2 + ...)*( x^0 + x^2 + x^4 + ...)
                                       // 第二次循环计算 ( x^0 + x^1 + x^2 + x^3 + 2 * x^4 + x^5 + x^6 + ...)*( x^0 + x^3 + x^6 + ...)

            for(j=0;j<=n;j++)          // j代表前一个括号里的每一项
                for(k=0;k+j<=n;k+=i)   // k代表后一个括号里的每一项
                {
                    c2[k+j] += c1[j];  // 用c2来保存乘得的结果 因为后一个括号中每一项的系数一定是1，所以可以直接 +=
                }
            for(j=0;j<=n;j++)          // 计算完毕后将两个括号合并成的大大括号中每一项的系数保存在c1中，c2置空
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf("%d
",c1[n]);
    }
}