Children’s Queue
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10040 Accepted Submission(s): 3198
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?
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#include<iostream> #include<iomanip> using namespace std; int main() { int n; int f[1001][101] = {0}; f[0][1]=1; f[1][1]=1; f[2][1]=2; f[3][1]=4; for(int i=4; i<1001; ++i) { for(int j=1;j<101;++j) { f[i][j]+=f[i-1][j]+f[i-2][j]+f[i-4][j]; f[i][j+1]+=f[i][j]/10000; f[i][j]%=10000; } } while(cin>>n) {int j=100; while(!f[n][j]) {j--;} cout<<f[n][j]; j--; for(;j>0;j--) printf("%04d", f[n][j]); cout<<endl; } return 0; }