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Number Sequence Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 141781 Accepted Submission(s): 34438
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
Author CHEN, Shunbao
Source |
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 using namespace std; 5 int f[10000]; 6 int main() 7 { 8 int a,b; 9 long long n; 10 while(cin>>a>>b>>n&&(a||b||n)){ 11 int i; 12 f[1]=1,f[2]=1; 13 for(i=3;i<10000;i++){ 14 f[i]=(a*f[i-1]+b*f[i-2])%7; 15 //如果有两个连着 =1，则后面的全部和前面相同，即出现了周期 16 //这时就没必要再进行下去了，跳出循环, i-2为周期 17 if(f[i]==1&&f[i-1]==1) break; 18 } 19 n=n%(i-2); 20 f[0]=f[i-2]; 21 cout<<f[n]<<endl; 22 } 23 return 0; 24 }
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原文地址：http://www.cnblogs.com/shenyw/p/5180869.html
