HDU-1005-Number Sequence【打表找规律】

题目链接：点击打开链接

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 158103    Accepted Submission(s): 38720

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

思路：数比较大，感觉无从下手。就打个表看看有没有规律啊 我个zz

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;int a,b,n;int f[510];int main(){while(~scanf("%d%d%d",&a,&b,&n)&&(a||b||n)){f[1]=1; f[2]=1;int cnt;bool flag=0;for(int i=3;i<=500;i++){f[i]=(a*f[i-1]+b*f[i-2])%7;//printf("--%d--",f[i]); 打表先找周期 if(f[i]==1&&f[i-1]==1) // 找到周期 {cnt=i;break;}if(f[i]==0&&f[i-1]==0) // 发现第三项往后全部为 0 {flag=1;break;}}if(flag){puts(n>=3?"0":"1");continue;}int t=cnt-2; // t 为周期 int ans=n%t;printf("%d\n",f[ans?ans:t]);}return 0;}