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
题意：有n个位置，男孩女孩排队，要求女孩至少要2个在一起。
思路：设f[n]表示，n个人的情况。情况一、在f[n-1]的情况后面加一个男孩；情况二、在f[n-2]的情况后面加两个女孩；情况三、在f[n-3]最后是男孩（等价于在f[n-4]个个数）的后面加三个女孩；
所以：f[n]=f[n-1]+f[n-2]+f[n-4];由于数据比较大，所以采用大数加法就可以了。
转载请注明出处：寻找&星空の孩子 题目链接：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 }