Problem Description
Here's your first sequence. This sequence has n numbers.
Suppose the first sequence is a0, a1, a2 an.
Then the second sequence is a0^a1, a1^a2, a2^a3 , an^a0.
Ask you what the value of the m-th sequence is
For example:
4 2
6 7 1 3
The first sequence is 67 1 3
The second sequence is 1 6 2 5
The third sequence is 7 4 7 4
and so on.
The example asks you for a second sequence. So sample output: 1 6 2 5
Ideas:
Xiangyu is huge. I can't think how to find the law. So give me an AC code, with a code for typing
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 100055;
ll a[N], b[N];
int main()
{
int n;
ll m;
while(cin >> n >> m)
{
for(int i = 0; i < n; i++)
cin >> a[i];
m--;
ll x = 1;
while(m)
{
if(m&1)
{
for(int i = 0; i < n; i++)
b[i] = a[i]^a[(i+x)%n];
memcpy(a, b, sizeof(b));
}
m >>= 1;
x <<= 1;
}
for(int i = 0; i < n; i++)
{
cout << a[i];
if(i == n-1) printf("\n");
else printf(" ");
}
}
return 0;
}
//#include<bits/stdc++.h>
//using namespace std;
//vector<int> a[10000], b[10000];
//int main()
//{
// int n, m, num;
// while(~scanf("%d %d", &n, &m))
// {
// for(int i = 0; i < n; i++)
// {
// scanf("%d", &num);
// a[i].push_back(num);
// b[i].push_back(num);
// }
// int flag = 0;
// for(;flag < 50; flag++)
// {
// for(int i = 0; i < n; i++)
// {
// for(int j = 0; j < a[(i+1)%n].size(); j++)
// {
// b[i].push_back(a[(i+1)%n][j]);
// }
// sort(b[i].begin(), b[i].end());
// int t = b[i].size();
// for(int j = 0; j < t - 1; j++)
// {
// if(b[i][j] == b[i][j+1])
// {
//
// b[i].erase(b[i].begin() + j);
// b[i].erase(b[i].begin() + j);
// j--;
// t -= 2;
// }
// }
// }
// for(int i = 0; i < n; i++)
// {
// a[i].assign(b[i].begin(), b[i].end());
// }
// printf("hang = %d ", flag+2);
// for(int i = 0; i < n; i++)
// {
// for(int j = 0; j < a[i].size(); j++)
// {
// printf("%d", a[i][j]);
// }
// printf(" ");
// }
// printf("\n");
// }
// }
// return 0;
//}