hdu Number Sequence

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Number Sequence

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 9   Accepted Submission(s) : 2

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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

周期，因为数据特别大，又是mod，所以应该有规律，但是这个只判断两个就断定是规律了，常规的递归超出内存。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int s[100000005];int main(){    int a,b,n;    int i,j;    s[1]=1;    s[2]=1;    while(scanf("%d%d%d",&a,&b,&n))    {        if(a==0&&b==0&&n==0)            break;        int ans=0;//记录周期        for(i=3;i<=n;i++)        {            s[i]=(a*s[i-1]+b*s[i-2])%7;            for(j=2;j<i;j++)            {                if(s[j-1]==s[i-1]&&s[j]==s[i])                {                    ans=i-j;                    break;                }            }            if(ans>0)                break;        }        if(ans>0)            s[n]=s[(n-j)%ans+j];        cout<<s[n]<<endl;    }    return 0;}

参考网址：

http://blog.csdn.net/hnust_xiehonghao/article/details/7933362