codeforces 450 B Jzzhu and Sequences

题意：给出f1=x,f2=y,f(i)=f(i-1)+f(i+1),求f(n)模上10e9+7

因为 可以求出通项公式：f(i)=f(i-1)-f(i-2)

然后

f1=x;

f2=y;

f3=y-x;

f4=-x;

f5=-y;

f6=-y+x;

f7=x; 发现是以6为循环的

还有注意下余数为正，就每次加上一个mod再模上mod

#include<iostream>  
#include<cstdio>  
#include<cstring> 
#include <cmath> 
#include<stack>
#include<vector>
#include<map> 
#include<set>
#include<queue> 
#include<algorithm>  
using namespace std;

typedef long long LL;
const int INF = (1<<30)-1;
const int mod=1000000007;
const int maxn=100005;

LL a[15];

int main(){
    LL x,y,n;
    cin>>x>>y;
    cin>>n;
    a[1]=(x+mod)%mod;
    a[2]=(y+mod)%mod;
    a[3]=(y-x+2*mod)%mod;
    a[4]=(-x+mod)%mod;
    a[5]=(-y+mod)%mod;
    a[0]=(-y+x+2*mod)%mod;
    cout<<a[n%6]<<"
";
    return 0;
}