HDUOJ Number Sequence找规律

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 137304    Accepted Submission(s): 33277

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

找规律，让循环自己找到它的循环周期，然后按照循环的周期去处理数据，减少运行时间。

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#include<stdio.h>int f[100000010];int main(){    int a,b,n,i,j,s;    f[1]=f[2]=1;    while(scanf("%d%d%d",&a,&b,&n)&&a+b+n)    {        s=0;        for(i=3; i<=n; i++)        {            f[i]=(f[i-1]*a+f[i-2]*b)%7;            for(j=2; j<i; j++)                if(f[i-1]==f[j-1]&&f[i]==f[j])                {                    s=i-j;                    break;                }            if(s>0)                break;        }        if(s>0)            f[n]=f[(n-j)%s+j];        printf("%d\n",f[n]);    }    return 0;}

大神的代码，实在看不懂啊！以后再好好研究吧

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#include<stdio.h>#include<iostream>using namespace std;int main(){    int a,b,n,i,j;    while(scanf("%d%d%d",&a,&b,&n)&&a+b+n)    {        int f[2]= {1,1};        for(i=0; i<(n%49-1)/2; i++)        {            f[0]=(a*f[1]+b*f[0])%7;            f[1]=(a*f[0]+b*f[1])%7;        }        if(n%2)            printf("%d\n",f[0]);        else            printf("%d\n",f[1]);    }    return 0;}