hdu1297 Children’s Queue （动态规划，n个人排成一列的方案数+高精加）

Children’s Queue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12548    Accepted Submission(s): 4110

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

 

Author

SmallBeer (CML)

 

Source

杭电ACM集训队训练赛（VIII）

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1465 1133 1480 1438 1466 

 

解析：f[i] 表示队列长度为 i 时的方案数，则：f[i]=f[i-1]+f[i-2]+f[i-4]

           1.f[i-1]：最后一个为男生，前 i-1 个合法

           2.f[i-2]：最后一个为女生，那么前一个必为女生，前 i-2 个合法

           3.f[i-4]：最后一个为女生，那么前一个必为女生，i-2为女生，i-3为男生，前 i-4 个合法，这种前 i-2 个不合法的情况下加上最后两个女生就合法了。

代码：

#include<cstdio>
#include<algorithm>
using namespace std;

typedef long long LL;
const LL maxn1=1000;
const LL maxn2=100;
const LL len=1e8;
LL f[maxn1+10][maxn2];

void add(LL a[],LL b[])
{
LL i,last=0;
a[0]=max(a[0],b[0]);
for(i=1;i<=a[0];i++)
{
a[i]+=b[i]+last;
last=a[i]/len,a[i]%=len;
}
if(last)a[++a[0]]=last;
}

int main()
{
LL n,i;
f[0][0]=1,f[0][1]=1;
f[1][0]=1,f[1][1]=1;
f[2][0]=1,f[2][1]=2;
f[3][0]=1,f[3][1]=4;
for(i=4;i<=maxn1;i++)
{
add(f[i],f[i-1]);
add(f[i],f[i-2]);
add(f[i],f[i-4]);
}
while(scanf("%I64d",&n)!=EOF)
{
printf("%I64d",f
[f
[0]]);
for(i=f
[0]-1;i>=1;i--)printf("%08I64d",f
[i]);
printf("\n");
}
return 0;
}