HDOJ1297

 

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5687    Accepted Submission(s): 1764

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

 

 

【大数的输出花费了好久时间。】

 

/*
题目大意：n个孩子站成一排，要求至少两个女孩站在一起。（男孩、女孩足够多），求有多少种站法。
假设n-1个孩子已经站好了，现在第n个孩子过去。
如果第n个孩子是男孩：则有f(n-1)种站法
如果第n个孩子是女孩，那么第n-1个也必须是女孩：
    如果前n-2个孩子是合法的，那么就有f(n-2)种站法
    如果前n-2个孩子是不合法的，但是加上最后两个女孩就是合法的了，那么有f(n-4)种站法。
n    f
0    1
1    1
2    2
3    4
4    7
-----
n可能为1000哦~所以要用大数。
*/
#include <stdio.h>
#include <string.h>
#define N 100
#define M 100000
long a[1001][N];
void Add(long *a,long *b,long *c)//c = a + b
{
    int i;
    long carry;
    carry = 0L;
    for (i=N-1;i>=0;i--)
    {
        c[i]=a[i]+b[i]+carry;
        carry = 0L;
        if (c[i]>=M)
        {
            c[i]-=M;
            carry=1L;
        }
    }
}
void output(long *a)
{
    int i=0;
    while(a[i] == 0L)
        i++;
    printf("%ld",a[i++]);//注意大数的输出格式。
    while (i!=N)
    {//注意a[i]小于M-1的位数的情况
        printf("%05ld",a[i++]);
    }
    printf("\n");
    
}
int main()
{
    int n,i;
    long temp[N];
    memset(a,0L,sizeof(a));
    a[0][N-1]=1;
    a[1][N-1]=1;
    a[2][N-1]=2;
    a[3][N-1]=4;
    for (i=4;i<=1000;i++)
    {
        if (i==50)
        {
            i=50;
        }
        memset(temp,0L,sizeof(temp));
        Add(a[i-1],a[i-2],temp);
        Add(temp,a[i-4],a[i]);
    }
    while (scanf("%d",&n)!=EOF)
    {
        output(a[n]);
    }
    return 0;
}