Children’s Queue（递推）
发布时间：2020-12-14 02:31:46 所属栏目：大数据 来源：网络整理
导读：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 headmas
Children’s QueueTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)
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?
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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)
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Output
For each test case,there is only one integer means the number of queue satisfied the headmaster’s needs.
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Sample Input
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Sample Output
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?? 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。
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? 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。
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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
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#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;
}
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