HDU2604Queuing
Problem Description
Queues and Priority Queues are data structures which are known to most computer scientists. The Queue occurs often in our daily life. There are many people lined up at the lunch time.
Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue as fmf or fff, we call it O-queue else it is a E-queue.
Your task is to calculate the number of E-queues mod M with length L by writing a program.
Now we define that ‘f’ is short for female and ‘m’ is short for male. If the queue’s length is L, then there are 2L numbers of queues. For example, if L = 2, then they are ff, mm, fm, mf . If there exists a subqueue as fmf or fff, we call it O-queue else it is a E-queue.
Your task is to calculate the number of E-queues mod M with length L by writing a program.
Input
Input a length L (0 <= L <= 10 6) and M.
Output
Output K mod M(1 <= M <= 30) where K is the number of E-queues with length L.
Sample Input
3 8
4 7
4 8
Sample Output
6
2
1
分析： 題目要求不含101 和111 的串
設f[n]為長度為n串符合條件的個數
則很明顯在長度為（n-1）並且符合條件的串後面加上一個0一定符合
如果在長度為n-1的串後面加上一個1的話我們得考慮n-1的串結尾的元素
如果是00的話我們可以看做是長度為n-3的串加上100
如果是10的話我們可以看做長度為n-4的串加上1100
因此f[n]=f[n-1]+f[n-3]+f[n-4];
方法一：直接遞推（耗時多可能超時）
#include <iostream>
#include <cstdio>
using namespace std;
int n,M;
int main()
{
int n,m;
int a[4];
while(~scanf("%d%d",&n,&M)){
a[0]=1;
a[1]=2;
a[2]=4;
a[3]=6;
for(int i=4;i<=n;i++){
a[4]=(a[0]+a[1]+a[3])%M;
a[0]=a[1];
a[1]=a[2];
a[2]=a[3];
a[3]=a[4];
}
if(n>=4)
cout<<a[4]%M<<endl;
else
cout<<a[n]%M<<endl;
}
return 0;
}方法二：（矩陣加速）
#include <iostream>
#include <cstdio>
using namespace std;
int n,M;
struct matrax
{
int m[4][4];
};
matrax A={
1,0,1,1,
1,0,0,0,
0,1,0,0,
0,0,1,0
};
matrax E;
void init()
{
for(int i=0;i<4;i++)
for(int j=0;j<4;j++)
E.m[i][j]=(i==j);
}
matrax multi(matrax a,matrax b)
{
matrax c;
for(int i=0;i<4;i++){
for(int j=0;j<4;j++){
c.m[i][j]=0;
for(int k=0;k<4;k++)
c.m[i][j]+=a.m[i][k]*b.m[k][j]%M;
c.m[i][j]%=M;
}
}
return c;
}
matrax power(matrax A,int k)
{
matrax ans=E,p=A;
while(k){
if(k&1){
ans=multi(ans,p);
k--;
}
k>>=1;
p=multi(p,p);
}
return ans;
}
int main()
{
init();
int a[4]={1,2,4,6};
while(cin>>n>>M){
matrax ans=power(A,n-3);
int x=0;
for(int i=0;i<4;i++)
x+=(ans.m[0][i]*a[4-i-1])%M;
/*for(int i=0;i<4;i++){
for(int j=0;j<4;j++)
cout<<ans.m[i][j]<<" ";
cout<<endl;
}*/
cout<<x%M<<endl;
}
return 0;
}