杭电1005

Number Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45754 Accepted Submission(s): 10082
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 
Recommend
JGShining

刚开始用递归做，提交的结果是Memory Limit Exceeded，上百度查了，应该是栈溢出，也就是内存限制超出。以后遇上超时或超内存的题目无法解决可以尝试找循环。本题就是一个循环题目。f（n）对7求模，所以他只有0,1,2,3,4,5,6这7中情况，但是给定的公式中有A,B对f(n)进行限制，所以有7*7=49种情况，49个数一个循环。

#include<stdio.h>int main(){    int a[100];int i,j,k,d,b,n;while (scanf ("%d%d%d",&d,&b,&n) != EOF){if (d==0&&b==0&&n==0)break;a[0]=1;a[1]=1;for (i=2; i<=48; i++){a[i]=(d*a[i-1]+b*a[i-2])%7;}n=n%48;printf("%d\n",a[n-1]);}return 0;}