hdu3706单调队列

很基础的单调队列

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-9;
const int N=100000+10,maxn=500+100,inf=0x3f3f3f;

ll q[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n,a,b;
    while(cin>>n>>a>>b){
        ll ans=1,s=0,t=0,res=1;
        for(int i=1;i<=n;i++)
        {
            res=(res*a)%b;
            while(s<t&&q[t-1]>res)t--;//单调递减队列
            q[t++]=res;
            ans=(ans*q[s])%b;
        }
        cout<<ans<<endl;
    }
    return 0;
}

能ac的错解

突然发现这代码是有问题的，没有考虑到s和t的间距问题，可能是数据太弱了，居然过了

所以重新改了一下，加一个id记录标号，如果s的标号小于i-a就说明差距大于a了，需要把s++

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-9;
const int N=100000+10,maxn=500+100,inf=0x3f3f3f;

ll q[N],id[N];
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n,a,b;
    while(cin>>n>>a>>b){
        ll ans=1,s=0,t=0,res=1;
        for(int i=1;i<=n;i++)
        {
            res=(res*a)%b;
            while(s<t&&q[t-1]>res)t--;//单调递减队列
            q[t++]=res;
            id[t-1]=i;
            if(id[s]<i-a)s++;
            ans=(ans*q[s])%b;
        }
        cout<<ans<<endl;
    }
    return 0;
}

正解