【OpenJudge9270】【Pku2440】【递推】DNA
DNA
【描述】
A kind of virus has attacked the X planet, and many lives are infected. After weeks of study, The CHO (Creature Healthy Organization) of X planet finally finds out that this kind of virus has two kind of very simple DNA, and can be represented by 101 and 111. Unfortunately, the lives on the planet also have DNA formed by 0s and 1s. If a creature's DNA contains the virus' DNA, it will be affected; otherwise it will not. Given an integer L, it is clear that there will be 2 ^ L different lives, of which the length of DNA is L. Your job is to find out in the 2 ^ L lives how many won't be affected?
【输入】
The input contains several test cases. For each test case it contains a positive integer L (1 <= L <= 10 ^ 6). The end of input is indicated by end-of-file.
【输出】
For each test case, output K mod 2005, here K is the number of lives that will not be affected.
【样例输入】
4
【样例输出】
9
【Solution】
根据题意，111、101是不合法的，所以我们
令f1[i]为长度为i且结尾三个数是000的所有合法01串数量，
令f2[i]为长度为i且结尾三个数是001的所有合法01串数量，
令f3[i]为长度为i且结尾三个数是010的所有合法01串数量，
令f4[i]为长度为i且结尾三个数是011的所有合法01串数量，
令f5[i]为长度为i且结尾三个数是100的所有合法01串数量，
令f6[i]为长度为i且结尾三个数是110的所有合法01串数量。
设目前长度为n。我们发现000的后两位、100的后两位和f1[n]的最后三位的前两位相等，而多出来的一位有两种情况0和1，选择符合f1[n]最后一位的就行，方案数不变，这样就把f1[n]的最后三位凑全了，f1[n]便可以从f1[n-1]和f5[n-1]转移来。所以f1[n]=f1[n-1]+f5[n-1]。
以此类推：
f2[n]=f1[n-1]+f5[n-1]
f3[n]·=f2[n-1]
f4[n]=f2[n-1]
f5[n]=f3[n-1]+f6[n-1]
f6[n]=f4[n-1]
令dp[n]=f1[n]+f2[n]+f3[n]+f4[n]+f5[n]+f6[n]。下面就是激(奇)动(技)人(淫)心(巧)的数学推导了：
dp[n]=f1[n]+f2[n]+f3[n]+f4[n]+f5[n]+f6[n]
=f1[n-1]+f5[n-1]+f1[n-1]+f5[n-1]+f2[n-1]+f2[n-1]+f3[n-1]+f6[n-1]+f4[n-1]
=dp[n-1]+f3[n-2]+f6[n-2]+f1[n-2]+f5[n-2]+f1[n-2]+f5[n-2]
=dp[n-1]+f2[n-3]+f4[n-3]+f1[n-3]+f5[n-3]+f3[n-3]+f6[n-3]+f1[n-3]+f5[n-3]+f3[n-3]+f6[n-3]
=dp[n-1]+dp[n-3]+f1[n-3]+f5[n-3]+f3[n-3]+f6[n-3]
=dp[n-1]+dp[n-3]+f1[n-4]+f5[n-4]+f3[n-4]+f6[n-4]+f2[n-4]+f4[n-4]
=dp[n-1]+dp[n-3]+dp[n-4]
1 #include <cstdio> 2 int N; 3 int dp[1000010]; 4 int main(){ 5 scanf("%d",&N); dp[1]=2; dp[2]=4 ;dp[3]=6; dp[4]=9; 6 for(int i=5;i<=N;++i) dp[i]=(dp[i-1]%2005+dp[i-3]%2005+dp[i-4]%2005)%2005; 7 printf("%d",dp[N]); 8 return 0; 9 }
