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 = (<<)-;
 const int mod=;
 const int maxn=;

 LL a[];

 int main(){
     LL x,y,n;
     cin>>x>>y;
     cin>>n;
     a[]=(x+mod)%mod;
     a[]=(y+mod)%mod;
     a[]=(y-x+*mod)%mod;
     a[]=(-x+mod)%mod;
     a[]=(-y+mod)%mod;
     a[]=(-y+x+*mod)%mod;
     cout<<a[n%]<<"\n";
     return ;
 }