hdu 1297 Children’s Queue（大数处理）

链接：hdu 1297

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

123

 

Sample Output

124

 

题意：n个人站成一队，女生不能单独站着，即如果有女生必须至少有两个女生相邻（可以没有女生），

     问有多少种排法

分析：情况一：（最后是以男生结尾的） 这时候是：f[n-1]+男；            

      情况二：（最后是以女生结尾的）           

               1.这时候有可能：f[n-2]+女+女                  

               2.还有可能：f[n-4]+男+女+女+女

     所以最后的递推式是：f[n]=f[n-1]+f[n-2]+f[n-4];

注：数较大，要用大数处理

AC代码：

#include<stdio.h>int a[1001][101]={0};int main(){    int i,j,n,m;    a[1][0]=1;    a[2][0]=2;    a[3][0]=4;    a[4][0]=7;    for(i=5;i<=1000;i++)        for(j=0;j<=100;j++){            a[i][j]+=a[i-1][j]+a[i-2][j]+a[i-4][j];            m=a[i][j]/10000;            a[i][j]%=10000;            a[i][j+1]+=m;        }    while(scanf("%d",&n)!=EOF){        for(i=100;i>=0;i--)            if(a[n][i]!=0)                break;        printf("%d",a[n][i]);        for(j=i-1;j>=0;j--)            printf("%04d",a[n][j]);        printf("\n");    }    return 0;}