HDU1005&&NEFU67 没有循环节

时间:2023-12-09 17:30:25

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147161    Accepted Submission(s):
35755

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
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#include <string.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
int main()
{
int a,b,n,i,data[];
while (cin>>a>>b>>n&&a&&b&&n)
{
data[]=;
data[]=;
for (i=;i<=;i++)
{
data[i]=((a*data[i-]%)+((b*data[i-]%%)))%;
if (data[i-]==&&data[i]==) //if(data[i-1])意思是只要是正数就行
break;
}
data[]=data[i-];
cout<<data[n%(i-)]<<endl;
}
return ;
}