Number Sequence（1005）

Number Sequence
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 105154    Accepted Submission(s): 25537




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




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# include <iostream>
# include <string>
using namespace std;




int fun(int a, int b, int n)
{
if (n==1)
return 1;
else if(n==2)
return 1;
else
return (a*fun(a, b, n-1)+b*fun(a, b, n-2))%7;
}


int main()
{
int a, b, n;
while(cin >> a >> b >> n)
    {
if(a == 0 && b == 0 && n == 0)
break;
else
cout << fun(a, b, n%49) << endl;
}
return 0;
}






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#include<stdio.h>
int main()
{
    int f[56],a,b,i,n;
    f[0]=1;f[1]=1;
    while(1)
    {
        scanf("%d%d%d",&a,&b,&n);
        if(!a && !b && !n)
            break;
        for(i=2;i<49;i++)
            f[i]=(a*f[i-1]+b*f[i-2])%7;
        printf("%d\n",f[(n-1)%49]);
    }
    return 0;
}




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#include <stdio.h>
#include<math.h>
int main()
{
    int a,b,i;
    long n,num[50];
    num[1]=num[2]=1;
    while(scanf("%d %d %ld",&a,&b,&n),a+b+n)
    {
     for(i=3;i<=48;i++)
     num[i%48]=(a*num[i-1]+b*num[i-2])%7;
     printf("%ld\n",num[n%48]);
    }
    return 0;
}