HDU 1005 Number Sequence
Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 154923 Accepted Submission(s): 37854
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).
1 2 10
0 0 0
5
#include<iostream>
using namespace std;
int main()
{
int cycle,i; //周期
int f[50];
int a,b,n;
f[1] = f[2] = 1;
while(cin>>a>>b>>n,a+b+n)
{
for( i = 3;i <= 49;i++)
{
f[i] = (a*f[i-1] + b*f[i-2])%7;
if(f[i] == f[i-1] && f[i] == 1) break;
}
cycle = i - 2;
n = n%cycle;
if(n==0)
cout<<f[cycle]<<endl;
else
cout<<f[n]<<endl;
}
return 0;
}
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