题目链接: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)
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
2 3 4
3 4 5
4 5 6
5 6 7
Sample Output
2
3
4
5
6
3
4
5
6
Author
WhereIsHeroFrom@HDU
Source
单调队列简单题;
卡内存,list过不了,deque,G++过的;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=1e7+,M=4e6+,inf=;
const ll INF=1e18+,mod=1e9+;
/// 数组大小
int d[N];
ll num[N];
int main()
{
ll n,a,b;
while(~scanf("%lld%lld%lld",&n,&a,&b))
{
ll ans=,base=a%b;
int s=,e=;
for(int i=;i<=n;i++,base=(base*a)%b)
{
num[i]=base;
while(s<e&&d[s]<i-a)s++;
while(e>s&&num[d[e]]>=num[i])e--;
d[++e]=i;
//cout<<num[d[s+1]]<<" "<<endl;
ans=ans*(num[d[s+]])%b;
}
printf("%lld\n",ans);
}
return ;
}