B - Number Sequence
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Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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).
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
真让我纠结死了这个题,一般这种大数据的题以前一个学长教过把结果存到一个数组里面,用时调用就可以了,可是这个题关系到三个变量,这种题一看数据很大就知道要找规律了,这个题以前做过,但是WA了,结合着百度,我想是不是规律只是由A,B影响的,所以变量只改为A,B后再看结果,发现所有的结果最小公共周期是48,也就是只需要先存48个结果,就可以算出大数据的结果了!由这个题,我明白了找规律是件多么需要仔细的一件事!
#include <stdio.h>
#include <stdlib.h>
int main()
{
long long n;
int a[1000];
int A,B,i;
while(scanf("%d%d%lld",&A,&B,&n)!=EOF)
{
a[1]=1;
a[2]=1;
if(A==0&&B==0&&n==0) break;
for(i=3; i<=49; i++)
{
a[i]=(A*a[i-1]+B*a[i-2])%7;
}
n=n%48;
printf("%d\n",a[n]);
}
return 0;
}