HDUOJ Children’s Queue
Children’s Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8835 Accepted Submission(s): 2813
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
大数。。 公式 f(x)=f(x-1)+f(x-2)-F(x-4);
代码:
1 #include<iostream>
2 #include<cstdio>
3 #define maxn 250
4 #define len 1000
5 using namespace std;
6 int a[len+1][maxn+1]={{1},{2},{4},{7}};
7 int main()
8 {
9 int i,j,n,s,c=0;
10 for(i=4;i<=len;i++)
11 {
12 for(c=j=0;j<=maxn;j++)
13 {
14 s=a[i-1][j]+a[i-2][j]+a[i-4][j]+c;
15 a[i][j]=s%10;
16 c=(s-a[i][j])/10;
17 }
18 }
19 while(cin>>n)
20 {
21 for(i=maxn;a[n-1][i]==0;i--);
22 for(j=i;j>=0;j--)
23 {
24 printf("%d",a[n-1][j]);
25 }
26 printf("n");
27 }
28 return 0;
29 }
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