Give you three integers n, A and B.
Then we define S i = A i mod B and T i = Min{ S k| i-A <= k <= i, k >= 1}
Your task is to calculate the product of T i (1 <= i <= n) mod B.
Input
Each line will contain three integers n(1 <= n <= 10 7),A and B(1 <= A, B <= 2 31-1).
Process to end of file.
Output
For each case, output the answer in a single line.
Sample Input
1 2 3 2 3 4 3 4 5 4 5 6 5 6 7
Sample Output
2 3 4 5 6
单调栈裸题
维护一个单调栈
栈内元素从小到大
栈内的所有的id元素也是从小到大
这样我们每次查询栈头元素就是我们要查询的最小值了
代码
#include <bits/stdc++.h>
using namespace std;
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
long long n, a, b;
while(~scanf("%lld%lld%lld",&n,&a,&b))
{
deque<pair<int,int> > q;//保证单调递增
long long ans=1;
long long tmp=1;
for (int i = 1; i <= n; i++)
{
tmp=(tmp*a)%b;
while(q.size()&&q.back().first>tmp) q.pop_back();
q.push_back(make_pair(tmp,i) );
while(i-q.front().second>a) q.pop_front();
ans=(ans*q.front().first)%b;
}
printf("%lld\n",ans);
}
}