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HDU 1028 Ignatius and the Princess III(母函数)

2019-11-06 07:31:02
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Ignatius and the PRincess III

Time Limit: 2000/1000 MS (java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20140    Accepted Submission(s): 14100Problem 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!" InputThe 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. OutputFor 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思路:题目说的很清楚,就是把一个整数给拆开,问共有多少种方案。果断母函数!!不懂母函数的可以先看看这篇文章点击打开链接AC代码:
#include <iostream>using namespace std;int a[1000],b[1000];int main(){    int num;    int i,j,k;    while(cin >> num)    {        for(i = 0; i <= num; i++)        {            a[i] = 1;            b[i] = 0;        }        for(i = 2; i <= num; i++)        {            for(j = 0; j <= num; ++j)                for(k = 0; k+j <= num; k+=i)                {                    b[j+k] += a[j];                }            for(j = 0; j <= num; ++j)            {                a[j] = b[j];                b[j] = 0;            }        }        cout << a[num] << endl;    }    return 0;}
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