acm之递归题目1

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

代码：

#include<stdio.h>//排队女生不能单独一个 int main(){    int s[1001][101]={0};    int i,n,j;    s[1][1]=1;    s[2][1]=2;    s[3][1]=4;    s[4][1]=7;    for(i=5;i<1001;i++){        for(j=1;j<101;j++){            s[i][j]+=s[i-1][j]+s[i-2][j]+s[i-4][j];            s[i][j+1]+=s[i][j]/10000;//保存进位             s[i][j]=s[i][j]%10000;        }    }    while(scanf("%d",&n)!=EOF){        int k=100;        while(!s[n][k--]);        printf("%d",s[n][k+1]);        for(;k>0;k--)            printf("%04d",s[n][k]);        printf("\n");    }    return 0;}