YT03-遞推求解課堂題目-1005 Children’s Queue-（6.7日-煙臺大學ACM預備隊解題報告）

kewlgrl發表於2015-06-13
ACM

Children’s Queue

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 77   Accepted Submission(s) : 29

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Problem Description

There are many students in PHT School. One day, the headmaster whose name is PigHeader wanted all students stand in a line. He prescribed that girl can not be in single. In other words, either no girl in the queue or more than one girl stands side by side. The case n=4 (n is the number of children) is like
FFFF, FFFM, MFFF, FFMM, MFFM, MMFF, MMMM
Here F stands for a girl and M stands for a boy. The total number of queue satisfied the headmaster’s needs is 7. Can you make a program to find the total number of queue with n children?

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of children (1<=n<=1000)

Output

For each test case, there is only one integer means the number of queue satisfied the headmaster’s needs.

Sample Input

1
2
3

Sample Output

1
2
4

Author

SmallBeer (CML)

Source

杭電ACM集訓隊訓練賽（VIII）

#include<stdio.h>
int main()
{
    int n;
    int f[1001][101] = {0};
    f[0][1] = 1;
    f[1][1] = 1;
    f[2][1] = 2;
    f[3][1] = 4;
    for(int i = 4; i < 1001; ++i)
    {
        for(int j = 1; j < 101; ++j)
        {
            f[i][j] += f[i - 1][j] + f[i - 2][j] + f[i - 4][j];    
            f[i][j + 1] += f[i][j] / 10000;    
            f[i][j] %= 10000;    
        }
    }
    while(scanf("%d", &n) != EOF)
    {
        int k = 100;
        while(!f[n][k--]);    
        printf("%d", f[n][k + 1]);    
        for(; k > 0; --k)
        {
            printf("%04d", f[n][k]);    
        }
        printf("\n");
    }
    return 0;
}


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