HDU 1005 Number Sequence

Posted 2020-06-13

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 ZJCPC2004

 

 

 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 }