HDU-3706 Second My Problem First

http://acm.hdu.edu.cn/showproblem.php?pid=3706

Second My Problem First

Time Limit: 12000/4000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1330 Accepted Submission(s):
503


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
Each line will contain three integers n(1 <= n <=
107),A and B(1 <= A, B <= 231-1).
Process to end
of file.

[align=left]Output[/align]
For each case, output the answer in a single
line.

[align=left]Sample Input[/align]

1 2 3
2 3 4
3 4 5

4 5 6

5 6 7

[align=left]Sample Output[/align]

2

3
4

5
6

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<cmath>
#define len 0x7fffffff
using namespace std;
struct node
{
__int64 num;
int p;
};
node q1[1000010];
//int p[1000010];
int main()
{
__int64  a,b;
int  n,i,t;
while(~scanf("%d%I64d%I64d",&n,&a,&b))
{
__int64 ti=1;
__int64  s=1;

int head=0,r=0;
for(i=1;i<=n;i++)
{
s=(s*a)%b;
while(head<r&&q1[r].num>=s) r--;
q1[++r].num=s;
q1[r].p=i;
while(q1[head+1].p<i-a)
head++;
ti=(ti*q1[head+1].num)%b;

}
printf("%I64d\n",ti);
}
return 0;
}