杭电 2100 Children’s Queue

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14884 Accepted Submission(s): 4940

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

递推
①如果是男生，那么就是 f（i - 1）；
②如果是女生，而f（i - 1）的最后一个有可能是男生，直接加女生不成立。所以就在f（i - 2）的后面直接放两个女生就满足了条件。 可是 还有一种特殊情况 ~~(>_<)~~，如果f（i - 2）的最后只有一女生（这种情况在f（i - 2）中不存在，但是在f（i）中可能存在，即连续三个女生。）这个时候就应该加上f（i - 4）（即f（i - 4）后面放一个男生 + 一个女生 + 两个女生）。
所以递推公式为 ：
f（n）= f（n - 1）+ f（n - 2）+ f（n - 4）
接下来 就是大数问题了 O(∩_∩)O

#include<iostream>#include<algorithm>#include<cstdio>using namespace std;int a[7500][800]={0};int main(){    int i,j,k,l;    a[1][0]=1;    a[0][0]=1;    a[2][0]=2;    a[3][0]=4;    k=0;    for(i=4;i<7500;i++)    {        for(j=0;j<=k;j++)        {            a[i][j]+=a[i-1][j]+a[i-2][j]+a[i-3][j];            if(a[i][j]>=10)            {                a[i][j+1]=a[i][j]/10;                a[i][j]%=10;            }        }        if(a[i][j]>=10)        {            k++;        }    }    while(scanf("%d",&l)!=EOF)    {        for(i=k;i>=0;i--)            if(a[l][i]!=0)                break;        for(;i>=0;i--)            printf("%d",a[l][i]);        printf("\n");    }    return 0;}