Children’s Queue（hdu1297+遞推），queuehdu1297

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12101 Accepted Submission(s): 3953

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   題意：有n個位置，男孩女孩排隊，要求女孩至少要2個在一起。   思路：設f[n]表示，n個人的情況。情況一、在f[n-1]的情況後面加一個男孩；情況二、在f[n-2]的情況後面加兩個女孩；情況三、在f[n-3]最後是男孩（等價於在f[n-4]個個數）的後面加三個女孩； 所以：f[n]=f[n-1]+f[n-2]+f[n-4];由於數據比較大，所以采用大數加法就可以了。   轉載請注明出處：尋找&星空の孩子 題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1297  

 1 #include<stdio.h>
 2 #include<string.h>
 3 int f[1005][105];
 4 void init()
 5 {
 6     memset(f,0,sizeof(f));
 7     f[0][1]=1;
 8     f[1][1]=1;
 9     f[2][1]=2;
10     f[3][1]=4;
11 
12     for(int i=4;i<=1000;i++)
13     {
14         int add=0;
15         for(int j=1;j<=100;j++)
16         {
17             f[i][j]=f[i-1][j]+f[i-2][j]+f[i-4][j]+add;
18             add=f[i][j]/10000;
19             f[i][j]%=10000;
20             if(add==0&&f[i][j]==0)break;
21         }
22     }
23 }
24 int main()
25 {
26     int n;
27     init();
28     while(scanf("%d",&n)!=EOF)
29     {
30         int k=100;
31         while(!f[n][k])k--;
32         printf("%d",f[n][k--]);
33         for(;k>0;k--)
34         {
35             printf("%04d",f[n][k]);
36         }
37         printf("\n");
38     }
39     return 0;
40 }