hdu 1297 Children’s Queue（大数处理）

最新推荐文章于 2020-05-21 20:40:35 发布

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探讨了一个有趣的问题：当女生不允许单独站在队伍中时，n个孩子可以有多少种不同的排队方式。通过递推公式解决了该问题，并给出了AC代码实现。

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

Sample Output

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;
}