hdu 1297 Children’s Queue (大数加法+递推)
2014-09-20 13:47
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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
Sample Output
分析过程:递推问题
设:F(n)表示n个人的合法队列,则:
按照最后一个人的性别分析,他要么是男,要么是女,所以可以分两大类讨论:
1、如果n个人的合法队列的最后一个人是男,则对前面n-1个人的队列没有任何限制,他只要站在最后即可,所以,这种情况一共有F(n-1);
2、如果n个人的合法队列的最后一个人是女,则要求队列的第n-1个人务必也是女生,这就是说,限定了最后两个人必须都是女生,这又可以分两种情况;
(1)如果队列的前n-2个人是合法的队列,则显然后面再加两个女生,也一定是合法的,这种情况有F(n-2);
(2)但是,难点在于,即使前面n-2个人不是合法的队列,加上两个女生也有可能是合法的,当然,这种长度为n-2的不合法队列,不合法的地方必须是尾巴,就是说,这里 说的长度是n-2的不合法串的形式必须是“F(n-4)+男+女”,这种情况一共有F(n-4).
所以,通过以上的分析,可以得到递推的通项公式: F(n)=F(n-1)+F(n-2)+F(n-4) (n>3)然后就是对n<=3 的一些特殊情况的处理了,显然:F(0)=1 (没有人也是合法的,这个可以特殊处理,就像0的阶乘定义为1一样) F(1)=1 F(2)=2 F(3)=4
#include <iostream>
#include <string.h>
using namespace std;
int F[1001][301];
void get()
{
memset(F,0,sizeof(F));
F[0][300]=1; F[1][300]=1;
F[2][300]=2; F[3][300]=4;
int i,j;
for(i=4;i<=1000;i++)
{
int sum=0;
for(j=300;j>=0;j--)
{
sum=sum+F[i-1][j]+F[i-2][j]+F[i-4][j];
F[i][j]=sum%10;
sum=sum/10;
}
}
}
int main()
{
get();
int N;
while(cin>>N)
{
int i,j;
for(i=0;F
[i]==0;i++)
j=i;
for(j=j+1;j<=300;j++)
cout<<F
[j];
cout<<endl;
}
return 0;
}
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
分析过程:递推问题
设:F(n)表示n个人的合法队列,则:
按照最后一个人的性别分析,他要么是男,要么是女,所以可以分两大类讨论:
1、如果n个人的合法队列的最后一个人是男,则对前面n-1个人的队列没有任何限制,他只要站在最后即可,所以,这种情况一共有F(n-1);
2、如果n个人的合法队列的最后一个人是女,则要求队列的第n-1个人务必也是女生,这就是说,限定了最后两个人必须都是女生,这又可以分两种情况;
(1)如果队列的前n-2个人是合法的队列,则显然后面再加两个女生,也一定是合法的,这种情况有F(n-2);
(2)但是,难点在于,即使前面n-2个人不是合法的队列,加上两个女生也有可能是合法的,当然,这种长度为n-2的不合法队列,不合法的地方必须是尾巴,就是说,这里 说的长度是n-2的不合法串的形式必须是“F(n-4)+男+女”,这种情况一共有F(n-4).
所以,通过以上的分析,可以得到递推的通项公式: F(n)=F(n-1)+F(n-2)+F(n-4) (n>3)然后就是对n<=3 的一些特殊情况的处理了,显然:F(0)=1 (没有人也是合法的,这个可以特殊处理,就像0的阶乘定义为1一样) F(1)=1 F(2)=2 F(3)=4
#include <iostream>
#include <string.h>
using namespace std;
int F[1001][301];
void get()
{
memset(F,0,sizeof(F));
F[0][300]=1; F[1][300]=1;
F[2][300]=2; F[3][300]=4;
int i,j;
for(i=4;i<=1000;i++)
{
int sum=0;
for(j=300;j>=0;j--)
{
sum=sum+F[i-1][j]+F[i-2][j]+F[i-4][j];
F[i][j]=sum%10;
sum=sum/10;
}
}
}
int main()
{
get();
int N;
while(cin>>N)
{
int i,j;
for(i=0;F
[i]==0;i++)
j=i;
for(j=j+1;j<=300;j++)
cout<<F
[j];
cout<<endl;
}
return 0;
}
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