hdoj 1297 Children’s Queue 【高精度】【递推】

题意:有n个人,每一个人可以是男孩也可以是女孩,要求每个女孩不能单独一个,也就是一个女孩的左右紧挨的位置至少要有一个女孩。问这样的队列有几个。

分析:设f(n)是n的排列的数目,这时候来一个人:

一:如果是男孩,那么f(n ) = f(n-1)

二:如果是女孩,如果前n-2是合法的,那么f(n) = f(n-2);如果前n-2不合法的,那么n-2队列的末尾两个同学肯定是男+女,那么再加上后来的女孩,又合法了,此时f(n) = f(n-4);

综上:f(n) = f(n-1)+f(n-2)+f(n-4);

又因为数据范围比较大,要用高精度。

代码:

#include <stdio.h>
#include <string.h>
#define M 1050
int a[M][M];
void table(){
    int i, j;
    memset(a, 0, sizeof(a));
    a[1][0] =1; a[2][0] = 2; a[3][0] = 4; a[4][0] = 7;
    for(i = 5; i < 1050; i ++){
        for(j = 0; j < M; j ++)
            a[i][j] = a[i-1][j]+a[i-2][j]+a[i-4][j];
        for(j = 0; j < M; j ++){
            if(a[i][j] >= 10){
                int temp = a[i][j]/10;
                a[i][j]%=10;
                a[i][j+1] += temp;
            }
        }
    }
}
int main(){
    table();
    int n, i, j;
    while(scanf("%d", &n) == 1){
        i = M-1;
        while(a[n][i] == 0) i --;
        printf("%d", a[n][i]);
        while((--i) >= 0) printf("%d", a[n][i]);
        puts("");
    }
    return 0;
} 

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1297

时间: 2024-11-07 03:40:32

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