HDU Children’s Queue

2024年1月19日 279次阅读

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 160 Accepted Submission(s): 102  

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

 
注意：
1、递归公式
1)m

2)mff

3)mfff

a:安全序列后加ff或者m，结果仍然安全。

b:不安全序列后加ff可使其安全，虽然mf加f也能得到安全序列，但与a情况重复。

故：公式a[n]=a[n-1]+a[n-2]+a[n-4];

2、大数，高精度问题。可以用二维数组模拟大数计算。n=1000时输出结果：

12748494904808148294446671041721884239818005733501580815621713101333980596197474

74433619974245291299822523591089179822154130383839594330018972951428262366519975

47955743099808702532134666561848656816661065088789701201682837073071502397487823

19037

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














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