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  • ACM学习历程—HDU1028 Ignatius and the Princess III(递推 || 母函数)

    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

    这是一个整数划分,母函数是构造了一个多项式的乘法,然后指数为n的一项的系数就是划分数。效率是n*n*n。

    递推稍微快一点,采用二位递推,p[i][j]表示i可以划分成j个数的划分个数。那么n的划分数就是sum(p[n][i])。

    对于p[i][j]:

    考虑最小的数,如果最小的数是1,就不再考虑这个1,那么就是p[i-1][j-1]。

    如果最小数不是1,那么可以对每个数都减一,那么就是p[i-j][j]。

    所以 p[i][j] = p[i-1][j-1]+(i-j >= 0 ? p[i-j][j] : 0);

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <cmath>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    #include <algorithm>
    #define LL long long
    
    using namespace std;
    
    int n, p[125][125];
    
    void work()
    {
        memset(p, 0, sizeof(p));
        p[1][1] = 1;
        for (int i = 2; i <= n; ++i)
            for (int j = 1; j <= n; ++j)
                p[i][j] = p[i-1][j-1]+(i-j >= 0 ? p[i-j][j] : 0);
        LL ans = 0;
        for (int i = 1; i <= n; ++i)
            ans += p[n][i];
        printf("%I64d
    ", ans);
    }
    
    int main()
    {
        //freopen("test.in", "r", stdin);
        while (scanf("%d", &n) != EOF)
            work();
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/andyqsmart/p/4742660.html
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