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(CF#257)B. Jzzhu and Sequences

2016-04-06 12:34 411 查看
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 -12


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;
}
}
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