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?
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。
#include<stdio.h> #include<string.h> int f[1005][105]; void init() { memset(f,sizeof(f)); f[0][1]=1; f[1][1]=1; f[2][1]=2; f[3][1]=4; for(int i=4;i<=1000;i++) { int add=0; for(int j=1;j<=100;j++) { f[i][j]=f[i-1][j]+f[i-2][j]+f[i-4][j]+add; add=f[i][j]/10000; f[i][j]%=10000; if(add==0&&f[i][j]==0)break; } } } int main() { int n; init(); while(scanf("%d",&n)!=EOF) { int k=100; while(!f[n][k])k--; printf("%d",f[n][k--]); for(;k>0;k--) { printf("%04d",f[n][k]); } printf("\n"); } return 0; }