Threeprime Numbers

1586. Threeprime Numbers

Time limit: 1.0 second

Memory limit: 64 MB

Rest at the sea is wonderful! However, programmer Pasha became awfully bored of lying on a beach in Turkey; so bored that he decided to count the quantity of three-digit prime numbers. This turned out
to be so interesting that he then started to study threeprime numbers. Pasha calls an integer a threeprime number if any three consecutive digits of this integer form a three-digit prime number. Pasha had already started working on the theory of the divine
origin of such numbers when some vandals poured water on Pasha and cried some incomprehensible words like “Sonnenstich!”, “Colpo di sole!”, and “Coup de soleil!”

You are to continue Pasha’s work and find out how often (or rare) threeprime numbers are.

Input

The input contains an integer n (3 ≤ n ≤ 10000).

Output

Output the quantity of n-digit threeprime numbers calculated modulo 109 + 9.

Sample

input	output
4	204

/**DP = =!**/
/**思路：用dp[j][k]表示位数时，最后两位是k，且满足题目要求的数字的个数**/
#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;
#define maxn 10010
#define mod 1000000009

bool prim[1010];
long long dp[maxn][100];
int n;

void deal()//求出100~1000内的素数，并初始化dp[3]
{
prim[2]=0;
for(int i=2;i<1000;i++)
if(!prim[i])
{
for(int k=2
;i*k<1000;k++)
prim[k*i]=1;
if(i>=100)
dp[3][i%100]++;
}
}

void read()
{
scanf("%d",&n);
}

void DP()
{
deal();
for(int i=4;i<=n;i++)
{
for(int j=10;j<100;j++)//枚举i位数字最后两位的值
for(int k=10;k<100;k++)//枚举i-1位数字最后两位的值
if(!prim[j%10+k*10]&&j/10==k%10)/**如果i位数字最后两位的值和i-1位最后一位的值构成素数，且i位数字倒数第二位数字等于i-1
位最后一位的数字**/
{
dp[i][j]+=dp[i-1][k];//状态转移
dp[i][j]%=mod;//注意取余
}
}
long long ans=0;
for(int i=0;i<100;i++)
{
ans+=dp
[i];
ans%=mod;
}
printf("%lld\n",ans);
}

int main()
{
read();
DP();
return 0;
}