Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 200250 Accepted Submission(s): 50305
Problem Description
A number sequence is defined as follows:
f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.
Given A, B, and n, you are to calculate the value of f(n).
Input
The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.
Output
For each test case, print the value of f(n) on a single line.
Sample Input
1 1 3
1 2 10
0 0 0
Sample Output
2
5
一开始以为避开递归的坑就行了,没想到Memory Limit Exceeded了。
原代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 #include<iomanip> 5 #include<cmath> 6 int F[100000001]; 7 using namespace std; 8 int main() 9 { 10 int A,B,n,i; 11 F[1]=1;F[2]=1; 12 while(cin>>A>>B>>n&&A!=0&&B!=0&&n!=0) 13 { 14 for(i=3;i<=n;i++) 15 F[i]=(A*F[i-1]+B*F[i-2])%7; 16 cout<<F[n]<<endl; 17 } 18 } 19
查了一下,说是48一个循环。可能是因为玄学。以后再去深究吧。
AC代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 #include<iomanip> 5 #include<cmath> 6 int F[48]; 7 using namespace std; 8 int main() 9 { 10 int A,B,n,i; 11 F[1]=1;F[2]=1; 12 while(cin>>A>>B>>n&&A!=0&&B!=0&&n!=0) 13 { 14 for(i=3;i<=47;i++) 15 F[i]=(A*F[i-1]+B*F[i-2])%7; 16 cout<<F[n%48]<<endl; 17 } 18 }