51NOD 1952 栈 【单调队列】

传送门

---

动态维护一个非严格单调递增队列即可
这样队列尾就是最大值
每个元素只可能进出队列各一次
O(n)

#include<stdio.h>#include<bits/stdc++.h>#define ll long long#define pii pair<int,int>#define pll pair<ll,ll>#define MEM(a,x) memset(a,x,sizeof(a))#define lowbit(x) ((x)&-(x))using namespace std;const int INF = 1e9+7;const int inf=INF;struct S{    deque<int>que;//原数据    deque<int>incrQue;//非严格单调递增队列    void init(){        que.clear();        incrQue.clear();    }    void push_back(int x){        que.push_back(x);        if(incrQue.empty()||x>=incrQue.back()){            incrQue.push_back(x);        }    }    void push_front(int x){        que.push_front(x);        while(!incrQue.empty()&&incrQue.front()<x){            incrQue.pop_front();        }        incrQue.push_front(x);    }    void pop_back(){        if(que.back()==incrQue.back()){            incrQue.pop_back();        }        que.pop_back();    }    int maxVal(){        return incrQue.back();    }    int size()const{        return que.size();    }}myStack;int main(){    //freopen("/home/lu/code/r.txt","r",stdin);    //freopen("/home/lu/code/w.txt","w",stdout);    int n,A,B,C,x0,a,b,MOD;    while(~scanf("%d%d%d%d%d%d%d%d",&n,&A,&B,&C,&x0,&a,&b,&MOD)){        myStack.init();        int ans=0;        int ta,tb;        for(int i=1;i<=n;++i){            int xi=((ll)x0*a+b)%MOD;            if(xi%(A+B+C)<A||myStack.size()<=1){                ta=0;                tb=xi;            }            else{                if(A<=xi%(A+B+C)&&xi%(A+B+C)<A+B){                    ta=1;                    tb=xi;                }                else{                    if(A+B<=xi%(A+B+C)){                        ta=2;                    }                }            }            switch (ta) {            case 0:                myStack.push_back(tb);                break;            case 1:                myStack.push_front(tb);                break;            case 2:                myStack.pop_back();            default:                break;            }            ans=(ans+myStack.maxVal())%inf;            x0=xi;        }        printf("%d\n",ans);    }    return 0;}