Jzzhu has invented a kind of sequences, they meet the following property:
You are given x and y, please calculate fn modulo 1000000007 (109 + 7).
Input
The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single integer n (1 ≤ n ≤ 2·109).
Output
Output a single integer representing fn modulo 1000000007 (109 + 7).
Sample test(s)
input
2 3 3
output
1
input
0 -1 2
output
1000000006
Note
In the first sample, f2 = f1 + f3, 3 = 2 + f3, f3 = 1.
In the second sample, f2 = - 1; - 1 modulo (109 + 7) equals (109 + 6).
本来9点的CF，今天有学姐来，讲到了9点半，这题最后没注意坑点，最后判的时候还wa了，掉了100分，蛋疼中
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1100;
const int M=1000000007;
int a[maxn];
int main()
{
int x,y,n;
while(cin>>x>>y>>n)
{
a[1]=x;
a[2]=y;
int len=0,t;
for(int i=3;;i++)
{
a[i]=a[i-1]-a[i-2];
if(a[i]==a[2]&&a[i-1]==a[1]&&i>=4)
{
len=i-2;
break;
}
if(i>=n)
break;
}
if(len)
{
// cout<<"len:"<<len<<endl;
t=(n-1)%len+1;
}
else
t=n;
if(a[t]>0)
cout<<a[t]%M<<endl;
else
{
while(a[t]<0)
a[t]+=M;
cout<<a[t]%M<<endl;
}
}
return 0;
}看了别人的想法，我的还是太狭隘了。我仅仅知道找规律，别人找的规律更详细。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int M=(1e9)+7;
int a[6];
int main()
{
int x,y,n;
while(cin>>x>>y>>n)
{
a[1]=(x+M)%M;
a[2]=(y+M)%M;
a[3]=(a[2]-a[1]+M)%M;
a[4]=(-x+M)%M;
a[5]=(-y+M)%M;
a[0]=(a[1]-a[2]+M)%M;
cout<<(a[n%6]+M)%M<<endl;
}
}