hdu 1005 Number Sequence

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 155586    Accepted Submission(s): 38044

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

 题目大意就是跟斐波那契数列差不多，a[i]=(A*a[i-1]+B*a[i-2])这道题的只能打下表，没有什么其他的方法暂时 要不然就会T掉（时间超限）；坑点有两个：

坑点1：如果是A和B是7的倍数的话n大于2所有的结果都为0；

坑点2；在写那个for循环的时候只能是i<=一个不会爆掉的常值（200都过了），取n的时候会T掉或者re掉（数组越界）不知道是怎么回事；

坑点3：输出的时候要判断是否是7的整数倍，如果是的话输出的是最后一位；

ac代码

#include<stdio.h>#include<string.h>#include<stdlib.h>#define N 100005int main(){    int a[N];    int k, n, i;    int A, B, ans, flag;    while(scanf("%d%d%d",&A,&B,&n))    {        if(A==0 && B==0 && n==0)            break;        a[0]=1;a[1]=1;flag=0;        for(i=2;i<N;i++)///就是这里的N取不对很容易爆掉，不能用n        {            a[i]=(A*a[i-1]+B*a[i-2])%7;            if(a[i]==1 && a[i-1]==1){ ans=i-1;break;}            if(a[i]==0 && a[i-1]==0){flag=1; break;}        }        if(flag==1) {printf("0\n");continue;}        if(i==n)    {printf("%d\n",a[n-1]);continue;}        if(n%ans==0){printf("%d\n",a[ans-1]);continue;}        printf("%d\n",a[n%ans-1]);    }    return 0;}