Problem - 1005_Number Sequence

Number Sequence

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

Total Submission(s): 109210    Accepted Submission(s): 26545

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

 

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#include<stdio.h>int f[55],y,z;void size_presenceOfCyclic(int a,int b){    int i,j;    f[1]=f[2]=1;    for(i=3;;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]){                y=j,z=i-1;                return;            }        }    }}int main(){    int a,b,n;    while(scanf("%d%d%d",&a,&b,&n)!=EOF&&(a!=0||b!=0||n!=0)){        size_presenceOfCyclic(a%7,b%7);        printf("%d\n",f[n<y?n:(n-y)%(z-y+1)+y]);    }    return 0;}