（HDOJ 1005）Number Sequence
时间：2011-04-13 来源：cpoint
Number Sequence
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).
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
AC code:
#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;
}
#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;
}
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