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;
}