Number Sequence F（n）的计算

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 174253    Accepted Submission(s): 43022

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

Recommend

JGShining

这个题要注意7*7=49种情况，这是模的规律。用最简单的方法会产生超时现象。。。

#include<stdio.h>int main () {int a,b,n;int xun[120];xun[1]=xun[2]=1,xun[0]=1;while(scanf("%d%d%d",&a,&b,&n)&&a|b|n){int shi,mo,k=0;for(int i=3;i<=n;++i){    xun[i]=(a*xun[i-1]+b*xun[i-2])%7;    for(int j=2;j<=i-1;++j)    {    if(xun[i]==xun[j]&&xun[i-1]==xun[j-1])    {    shi=j;    mo=i;    k=1;break;}}if(k==1)break;}if(k==1){printf("%d\n",xun[shi+(n-mo)%(mo-shi)]);}elseprintf("%d\n",xun[n]);}}