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 likeFFFF, 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
123
Sample Output
124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，
Children’s Queue（递推）
，电脑资料
《Children’s Queue（递推）》(https://www.unjs.com)。 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#include<stdio.h>#include<string.h>int f[1005][105];void init(){ memset(f,0,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;}</string.h></stdio.h>