hdu1105

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 153645    Accepted Submission(s): 37479

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   |   We have carefully selected several similar problems for you:  1008 1004 1021 1019 1002

解题思路：由于题目给出的数据1 <= n <= 100,000,000，数据太大，若按照题目意思进行递归，定会超时，所以只能寻找规律。
解题心得：在比赛的时候，就想着这题目一点规律也没有，也不好证明，就一直在纠结，这种题目就不应该出现在acm中，简直折磨人。到后面自己心平气和的坐下来，慢慢的思考这个问题时，发现自己还是太急躁了，我可以用递归的那种方法写出程序，再打印出结果，也不用自己算，慢慢的找规律。还是要少一点抱怨，多用一点脑子。
用于寻找规律的代码

#include <iostream>#include<cstdio>using namespace std;int a,b,n,c[100000005];int main(){    while(scanf("%d%d%d",&a,&b,&n)&&a!=0&&b!=0&&n!=0)    {        c[1]=1;        c[2]=1;        printf("1 1 ");        for(int i=3;i<=n;i++)        {            c[i]=(a*c[i-1])%7+(b*c[i-2])%7;            printf("%d ",c[i]);        }        printf("\n");    }    return 0;}

各种测试数据的结果：

ac代码：

#include <iostream>#include<cstdio>using namespace std;int a,b,n,i,c[10000];int main(){    c[1]=1;    c[2]=1;    while(scanf("%d%d%d",&a,&b,&n)&&a&&b&&n)    {        for(i=3;i<10000;i++)        {            c[i]=(a*c[i-1]+b*c[i-2])%7;            if(c[i]==1&&c[i-1]==1)                break;        }        n=n%(i-2);        c[0]=c[i-2];        printf("%d\n",c[n]);    }    return 0;}