B解题报告
算是规律题吧，，，x y z -x -y -z
注意的是假设数是小于0,要先对负数求模再加模再求模,不能直接加mod，可能还是负数
给我的戳代码跪了，，。
#include <iostream>
#include <cstring>
#include <cstdio> using namespace std;
long long x,y,z;
long long n;
int main()
{
cin>>x>>y;
cin>>n;
z=y-x;
long long t;
t=(n-1)/3;
if(t%2==0)
{
n%=3;
if(n==0)
{
if(z>=0)
cout<<z%1000000007;
else cout<<(z%1000000007+1000000007)%1000000007;
}
else if(n==1)
{
if(x>=0)
cout<<x%1000000007;
else cout<<(x%1000000007+1000000007)%1000000007;
}
else
{
if(y>=0)
cout<<y%1000000007;
else cout<<(y%1000000007+1000000007)%1000000007;
}
}
else
{
x=-x;
y=-y;
z=-z;
n%=3;if(n==0)
{
if(z>=0)
cout<<z%1000000007;
else cout<<(z%1000000007+1000000007)%1000000007;
}
else if(n==1)
{
if(x>=0)
cout<<x%1000000007;
else cout<<(x%1000000007+1000000007)%1000000007;
}
else
{
if(y>=0)
cout<<y%1000000007;
else cout<<(y%1000000007+1000000007)%1000000007;
}
}
return 0;
}
Jzzhu and Sequences
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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).