杭电1005Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 126495    Accepted Submission(s): 30758

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 31 2 100 0 0

 

Sample Output

25

 

#include<stdio.h>int main(){int n,x,y,z,i;int a[100];a[1]=1;a[2]=1;while(~scanf("%d %d %d",&x,&y,&z)){if(x==0&&y==0&&z==0)return 0;for(i=3;i<=49;i++) {a[i]=(x*a[i-1]+y*a[i-2])%7;}printf("%d\n",a[z%49]);}return 0;}

注意：int的数值范围有限，当数值z大时会有数值溢出，又因为对7取余，所以以49为周期取z%49的数值代替z；
for循环要i<=49结束
根据周期，可以快速获得结果，且周期必定小于49.
   周期的原因：如果<f(x),f(x+1)>的值与<f(x+T),f(x+1+T)>一致，那么以后的序列就完全一致了。
   周期小于49的原因： 由于mod 7，所以<f(x),f(x+1)>的可能为49个，那么50个这样的<f(x),f(x+1)>必定有两个相同（鸽笼原理），所以最终可以确定周期小于49.
2、需要注意的是：并不是整个序列是有周期的，而是其中去掉头部一部分以后是有周期的！！！举个特殊的例子A=7，B=7,那么{f(x)}为{1,1,0,0,0,0，...}，其在去掉头部的两个1后是有周期的，所以应该求出这个头部。类似的例子还有{A=2，B=7},{A=3,B=7}...