CodeForces 450B Jzzhu and Sequences 费波纳茨数列+找规律+负数MOD

题目：Click here

题意：给定数列满足求f(n)mod(1e9+7)。

分析：规律题，找规律，特别注意负数取mod。

#include <iostream>
#include <cstdio>
#include <cstring> 
#include <algorithm>
#include <cmath>
using namespace std;
const int M = 1e5+5;
const int MOD = 1e9+7;

int x, y, n;
int a[6];
int main()  {
    while( ~scanf("%d%d%d", &x, &y, &n ) )  {
        a[0] = x;
        a[1] = y;
        for( int i=0; i<6; i++ )    {
            if( i >= 2 )    
                a[i] = a[i-1] - a[i-2];
            while( a[i] < 0 )   {   //消除负数取mod的影响，将其转换成正数
                a[i] += MOD;
            }
            a[i] %= MOD;
        }
        printf("%d\n", a[(n-1)%6] );
    }
    return 0;
}