Number Sequence
http://acm.hdu.edu.cn/showproblem.php?pid=1005
找循环节。。
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1 #include <stdio.h> 2 #include <string.h> 3 const int N=52; 4 int f[N]; 5 int main() 6 { 7 int a,b,n; 8 while(~scanf("%d%d%d",&a,&b,&n)) 9 { 10 int s,e; 11 if (a==0&&b==0&&n==0) 12 break; 13 f[0] = 1; 14 f[1] = 1; 15 f[2] = 1; 16 for (int i = 3; i <= n; i++) 17 { 18 f[i] = (a*f[i-1]+b*f[i-2])%7; 19 for (int j = 2; j < i; j++) 20 { 21 if (f[i]==f[j]&&f[i-1]==f[j-1]) 22 { 23 s = j; 24 e = i; 25 goto RL; 26 } 27 } 28 } 29 printf("%d\n",f[n]); 30 continue; 31 RL: 32 printf("%d\n",f[s+(n-e)%(e-s)]); 33 } 34 return 0; 35 }
用java写了一下,果断的超时了。。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 import java.math.BigInteger; 2 import java.util.Scanner; 3 4 public class Main { 5 6 public static void main(String[] args) { 7 8 Scanner cin = new Scanner(System.in); 9 BigInteger M = BigInteger.valueOf(7); 10 while (cin.hasNext()) { 11 BigInteger a = cin.nextBigInteger(); 12 BigInteger b = cin.nextBigInteger(); 13 BigInteger n = cin.nextBigInteger(); 14 int s1 = a.compareTo(BigInteger.ZERO); 15 int s2 = b.compareTo(BigInteger.ZERO); 16 int s3 = n.compareTo(BigInteger.ZERO); 17 if (s1 == 0 && s2 == 0 && s3 == 0) 18 break; 19 BigInteger f1 = BigInteger.ONE; 20 BigInteger f2 = BigInteger.ONE; 21 BigInteger cnt = BigInteger.valueOf(2); 22 BigInteger f; 23 while (true) { 24 f = a.multiply(f2).add(b.multiply(f1)).mod(M); 25 f1 = f2; 26 f2 = f; 27 cnt = cnt.add(BigInteger.ONE); 28 s1 = n.compareTo(cnt); 29 if (s1 == 0) 30 break; 31 } 32 System.out.println(f); 33 34 } 35 } 36 37 } 38