# Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 169920 Accepted Submission(s): 41910
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*

``````
int f(int n){
if(n==1||n==2)
return 1;
return (A*f(n-1)+B*f(n-2))%7;
}
``````

`````` int a,b,c;
a=1;b=1;
if(n==1||n==2)
printf("1\n");
else{
for(int i=2;i<n;++i){
c=(A*b+B*a)%7;
a=b;
b=c;
}``````

1.n的数值很大，这类数值很大的问题一般都有规律，找出循环节（周期）是关键；
2.找规律，这道题是从 f(1) = 1 和 f(2) = 1 开始，然后依次模7，可知 f(n) 只有7种情况，所以两数相邻只有7*7=49种；
3.所以从 f(1) 到 f(49) 必会出现相邻两个 f(m-1) = 1 , f(m) = 1，所以 f(n) 为周期函数，`49为其一个周期`

AC 代码：

``````#include<cstdio>
int A,B;
int main(){
int n;
while(~scanf("%d %d %d",&A,&B,&n)&&A&&B&&n){
int a,b,c;
a=1;b=1;
n%=49;
if(n==1||n==2)
printf("1\n");
else{
for(int i=2;i<n;++i){
c=(A*b+B*a)%7;
a=b;
b=c;
}
printf("%d\n",c);
}
}
}``````

HDU 1005 Number Sequence【序列号】