Number Sequence
2015-10-30 08:50
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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
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
#include<iostream> #include<cstdio> using namespace std; int main(){ int f[50]; int a,b,i,n,num; while(cin >> a >> b >> n){ if(a==0 && b==0 && n==0) break; f[1]=1;f[2]=1; for(i=3;i<50;i++){ f[i]=(a*f[i-1]+b*f[i-2])%7; if(f[i]==1 && f[i-1]==1) break; } num=i; num=num-2; n=n%num; if(n==0) printf("%d\n",f[num]); else printf("%d\n",f ); } return 0; }
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