杭电 ACM Runtime Error(ACCESS_VIOLATION) 求解

Question

Number Sequence
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 22 Accepted Submission(s) : 10
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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

#include<stdio.h>
void main()
{
int a,b,n,i,k[49],t;
while(scanf("%d %d %d",&a,&b,&n)!=EOF&&(a!=0||b!=0||n!=0))
{
k[0]=1;
k[1]=1;
for(i=2;i<49;i++)
k[i]=(b*k[i-2]+a*k[i-1])%7;
for(i=2;i<49;i++)
if(k[i-1]==1&&k[i]==1)
t=i-1;
printf("%d\n",k[(n-1)%t]);
}
}
可是最小公倍数为啥是1008……

Answer 1

#include<stdio.h>
void main()
{
int a,b,n,i,k[2005],t;
while(scanf("%d %d %d",&a,&b,&n)!=EOF&&(a!=0||b!=0||n!=0))
{
k[0]=1;
k[1]=1;
for(i=2;i<=1008;i++)
k[i]=(b*k[i-2]+a*k[i-1])%7;
for(i=2;i<=1008;i++)
if(k[i-1]==1&&k[i]==1)
t=i-1;
printf("%d\n",k[(n-1)%t]);
}
}

Answer 2

周期的最小公倍数为1008