HDU Number Sequence

Number Sequence

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

Total Submission(s): 930 Accepted Submission(s): 367

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 3
1 2 10
0 0 0

 

Sample Output

2
5

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

 

Recommend

JGShining

 

 

对于公式 f[n] = A * f[n-1] + B * f[n-2]; 后者只有7 * 7 = 49 种可能，为什么这么说，因为对于f[n-1] 或者 f[n-2] 的取值只有 0,1,2,3,4,5,6 这7个数，A，B又是固定的，所以就只有49种可能值了。由该关系式得知每一项只与前两项发生关系，所以当连续的两项在前面出现过循环节出现了，注意循环节并不一定会是开始的 1，1 。 又因为一组测试数据中f[n]只有49中可能的答案，最坏的情况是所有的情况都遇到了，那么那也会在50次运算中产生循环节。找到循环节后，就可以轻松解决了。

 

 

#include<stdio.h>

int a,b,n;
int f[60];

int main(){
    int i,j;
    int beg,end;
    f[0]=f[1]=f[2]=1;
    while(scanf("%d%d%d",&a,&b,&n)){
        if(a==0 && b==0 && n==0)
            break;
        int flag=0;
        for(i=3;i<=n;i++){
            f[i]=(a*f[i-1]+b*f[i-2])%7;
            for(j=2;j<i;j++){
                if(f[i]==f[j] && f[i-1]==f[j-1]){
                    beg=j,end=i;
                    flag=1;
                    break;
                }
            }
            if(flag)
                break;
        }
        if(flag)
            printf("%d\n",f[beg+(n-end)%(end-beg)]);
        else
            printf("%d\n",f[n]);
    }
    return 0;
}