杭电ACM1297-Children’s Queue

http://acm.hdu.edu.cn/showproblem.php?pid=1297

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4958    Accepted Submission(s): 1518

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

杭电ACM集训队训练赛（VIII）

 

 

Recommend

lcy

 

我的理解

 

#include <stdio.h>
const int maxn=1000;
int m[1002][1000]={{0},{1,1},{1,1}},
    f[1002][1000]={{0},{1,0},{1,1}};
void add(int a[],int b[],int c[])
{
    int i;
    for (i=0;i<=a[0];i++) c[i]=a[i];
    if (b[0]>c[0]) c[0]=b[0];
    for (i=1;i<=c[0];i++)
    {
        c[i]+=b[i];
        if (c[i]>9)
        {
            c[i]-=10;
            c[i+1]++;
        }
        if (c[c[0]+1]>0) c[0]++;
    }
}
int main()
{
    int i,n;
    for (i=3;i<=maxn+1;i++)
    {
        add(m[i-1],f[i-1],m[i]);
        add(f[i-1],m[i-2],f[i]);
    }
    while (scanf("%d",&n)!=EOF)
    {
        for (i=m[n+1][0];i>0;i--) printf("%d",m[n+1][i]);
        printf("\n");
    }
}