Problem Description
Give you three integers n, A and B.
Then we define Si = Ai mod B and Ti = Min{ Sk | i-A <= k <= i, k >= 1}
Your task is to calculate the product of Ti (1 <= i <= n) mod B.
Then we define Si = Ai mod B and Ti = Min{ Sk | i-A <= k <= i, k >= 1}
Your task is to calculate the product of Ti (1 <= i <= n) mod B.
Input
Each line will contain three integers n(1 <= n <= 107),A and B(1 <= A, B <= 231-1).
Process to end of file.
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
这题可以用单调队列做,存储每次的时间和大小。
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
#define maxn 10000005
#define ll __int64
ll q[maxn][2];
int main()
{
ll n,m,A,B,ans,front,rear,time,sum,timenow;
int i,j;
while(scanf("%I64d%I64d%I64d",&n,&A,&B)!=EOF)
{
front=1;rear=0;ans=1;sum=1;
for(i=1;i<=n;i++){
ans=(ans*A)%B;
if(i-A<1)time=1;
else time=i-A;
while(front<=rear && ans<=q[rear][0]){
rear--;
}
rear++;
q[rear][0]=ans;q[rear][1]=i;
while(front<=rear && q[front][1]<time)front++;
sum=(sum*q[front][0])%B;
}
printf("%I64d
",sum%B);
}
return 0;
}