4293: [PA2015]Siano

去成都考试的题目的原题？
思路是一眼内容，就是代码不好写.
直接考虑用线段树维护一下区间草的信息即可…
c++代码如下：

/*
code by Tgotp
*/
#include<iostream>
#include<cctype>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#define lowbit(x) (x&-x)
#define eps 1e-8
#define rep(i,x,y) for(register int i = x ; i <= y; ++ i)
#define repd(i,x,y) for(register int i = x ; i >= y; -- i)
using namespace std;
typedef long long ll;
template<typename T>inline bool chkmax(T&x,T y) { return x < y ? x = y,1 : 0; }
template<typename T>inline bool chkmin(T&x,T y){ return x > y ? x = y,1 : 0; }
template<typename T>inline void read(T&x)
{
    x = 0;char c;int sign = 1;
    do { c = getchar(); if(c == '-') sign = -1; }while(!isdigit(c));
    do { x = x * 10 + c - '0'; c = getchar(); }while(isdigit(c));
    x *= sign;
}

inline void init(string name)
{
    string in = name + ".in",out = name + ".out";
    freopen(in.c_str(),"r",stdin);
    freopen(out.c_str(),"w",stdout);
}

const int N = 5e5+7,inf = 1e9+7;
int a[N],n,m;
struct segment_Tree
{
    ll lzy1[N<<2],lzy2[N<<2],sum[N<<2],tot[N<<2];
    ll mi1[N<<2],mx1[N<<2];
    ll mx2[N<<2],mi2[N<<2];

    void build(int id,int l,int r)
    {
        lzy1[id] = -1;
        if(l == r)
        {
            mi1[id] = mx1[id] = a[l];
            tot[id] = a[l];
            return;
        }

        int mid = l + r >> 1;
        build(id<<1,l,mid); build(id<<1|1,mid + 1,r);
        mi1[id] = min(mi1[id<<1],mi1[id<<1|1]);
        mx1[id] = max(mx1[id<<1],mx1[id<<1|1]);
        tot[id] = tot[id<<1] + tot[id<<1|1];
    }

    inline void put_down(int id,int l,int r)
    {
        if(l > r) return;
        if(~lzy1[id])
        {
            if(l != r) lzy1[id << 1] = lzy1[id<<1|1] = lzy1[id];
            mx2[id] = mi2[id] = lzy1[id];
            sum[id] = lzy1[id] * (r - l + 1);
            lzy1[id] = -1;lzy2[id] = 0;
        }
        else if(lzy2[id])
        {
            if(l != r)
            {
                if(~lzy1[id<<1]) put_down(id<<1,l,l+r>>1);
                if(~lzy1[id<<1|1]) put_down(id<<1|1,(l+r>>1) + 1,r);
                lzy2[id<<1] += lzy2[id];
                lzy2[id<<1|1] += lzy2[id];
            }
            mx2[id] += mx1[id] * lzy2[id];
            mi2[id] += mi1[id] * lzy2[id];
            sum[id] += tot[id] * lzy2[id];
            lzy2[id] = 0;
        }
    }

    void modify(int id,int l,int r,ll num,ll k)
    {
        if(mi1[id] * num + mi2[id] >= k)
        {
            mi2[id] = k; mx2[id] = k;
            sum[id] = k * (r - l + 1);
            if(l != r)
            {
                lzy1[id<<1] = k;
                lzy1[id<<1|1] = k;
             } 
            return ;
        }
        if(mx1[id] * num + mx2[id] <= k)
        {
            if(l!= r)
            {
                if(~lzy1[id<<1]) put_down(id<<1,l,l+r>>1);
                if(~lzy1[id<<1|1]) put_down(id<<1|1,(l+r>>1) + 1,r);
            }
            mi2[id] += num * mi1[id];
            mx2[id] += num * mx1[id];
            sum[id] += num * tot[id];
            if(l != r)
            {
                lzy2[id<<1] += num;
                lzy2[id<<1|1] += num;
            }
            return ;
        }
        if(l == r)
        {
            mi2[id] = mx2[id] = min(1ll*k,mi2[id] + mi1[id] * num);
            sum[id] = mi2[id];
            return;
        }
        int mid = l + r >> 1;
        put_down(id<<1,l,mid); put_down(id<<1|1,mid + 1,r);
        modify(id<<1,l,mid,num,k); modify(id<<1|1,mid + 1,r,num,k);

        mi2[id] = min(mi2[id<<1],mi2[id<<1|1]);
        mx2[id] = max(mx2[id<<1],mx2[id<<1|1]);
        sum[id] = sum[id<<1] + sum[id<<1|1];
    }

    ll query(int id,int l,int r,ll num,ll k)
    {
        if(mx1[id] * num + mx2[id] < k) return 0;
        if(mi1[id] * num + mi2[id] >= k) return sum[id] + tot[id] * num - (r - l + 1) * k;
        if(l == r) return max(1ll*0,mx2[id] + mx1[id] * num - k);
        int mid = l + r >> 1;
        put_down(id<<1,l,mid); put_down(id<<1|1,mid + 1,r);
        return query(id<<1,l,mid,num,k) + query(id<<1|1,mid + 1,r,num,k);
    }

}seg;

int main()
{
//  init("grass");

    read(n); read(m);
    rep(i,1,n) read(a[i]);
    sort(a + 1,a + 1 + n);
    seg.build(1,1,n);
    ll d,b,lst=0;
    while(m--)
    {
        read(d); read(b);
        printf("%lld\n",seg.query(1,1,n,d - lst,b));
        seg.modify(1,1,n,d - lst,b);
        lst = d;
    }

    return 0;
}