杭电ACM1005 Number Sequence

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

我的代码：
#include
int main(void)
{
__int64 a[100000];
int A,B,n,i;
while(scanf("%d%d%d",&A,&B,&n)!=EOF&&!(A==0&&B==0&&n==0))
{
a[1]=1;a[2]=1;
for(i=3;i<=n;i++)
a[i] = (A * a[i - 1] + B * a[i - 2])%7;
printf("%I64d\n",a[n]);
}
}

#include

int main(void) {
int a1 = 1,a2 = 1,an;
int A,B,n,i;
while(1) {
scanf("%d%d%d",&A,&B,&n);
if(A == 0 && B == 0 && n == 0) break;
a1 = 1;
a2 = 1;
for(i = 3; i <= n; ++i) {
an = (A * a2 + B * a1) % 7;
a1 = a2;
a2 = an;
}
printf("%d\n",an);
}
return 0;
}

// 每个数据都走那么多肯定是tle啦，这题是有周期的。 // 下面是我的ac代码 #include int main() { int a,b,n; int f[201]; int i; scanf("%d %d %d",&a,&b,&n); while (a!=0||b!=0||n!=0) { f[1]=1; f[2]=1; if (n>=3) { for(i=3;i<200;i++) { f[i]=(a*f[i-1]+b*f[i-2])%7; if ((f[i-1]==1)&&(f[i]==1)) { break; } } i=i-2; n=n%i; if (n==0) { n=i; } printf("%d\n",f[n]); } else { printf("1\n"); } scanf("%d %d %d",&a,&b,&n); } return 0; }