CodeForces-992E King-Shamans

传送门：CodeForces-992E

思路分析

用线段树维护$a_i-sum_{i-1}$，修改的时候区间修改，记录区间最大值和最小值
要找等于0的位置，就看区间是否可能包含0，减掉不可能的区间，复杂度还是很玄学的（跑挺快的）

AC代码

#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define int long long
namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 2e5+10;

struct node{
    int l,r,mid,len;
    ll ma,mi,lazy;
}tree[N<<2];

int a[N];
ll ss[N];

void pp(int x){
    tree[x].mi=min(tree[ls].mi,tree[rs].mi);
    tree[x].ma=max(tree[ls].ma,tree[rs].ma);
}

void build(int x,int l,int r){
    tree[x]={l,r,l+r>>1,r-l+1,-INF,INF,0};
    if(l==r){
        tree[x].ma=tree[x].mi=a[l]-ss[l-1];
        return ;
    }
    int mid=tree[x].mid;
    build(ls,l,mid);
    build(rs,mid+1,r);
    pp(x);
}

void pd(int x){
    if(tree[x].lazy){
        int s=tree[x].lazy;
        tree[rs].lazy+=s;
        tree[ls].lazy+=s;
        tree[rs].ma+=s;
        tree[ls].ma+=s;
        tree[rs].mi+=s;
        tree[ls].mi+=s;    
        tree[x].lazy=0;
    }
}

void modify(int x,int l,int r,int k){
    if(l<=tree[x].l && tree[x].r<=r){
        tree[x].ma+=k;
        tree[x].mi+=k;
        tree[x].lazy+=k;
        return ;
    }
    pd(x);
    int mid=tree[x].mid;
    if(l<=mid) modify(ls,l,r,k);
    if(r>mid) modify(rs,l,r,k);
    pp(x);
}

int ans;

void query(int x){
    if(ans!=-1) return ;
    if(tree[x].l==tree[x].r){
        if(tree[x].ma==0) ans=tree[x].l;
        return ;
    }
    pd(x);
    if(tree[ls].ma>=0 && tree[ls].mi<=0) query(ls); 
    if(tree[rs].ma>=0 && tree[rs].mi<=0) query(rs);
    pp(x);
}

signed main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int n,m;
    read(n,m);
    for(int i=1;i<=n;i++){
        read(a[i]);
        ss[i]=ss[i-1]+a[i];
    }
    build(1,1,n);
    while(m--){
        ans=-1;
        int p,x;
        read(p,x);
        int k=x-a[p];
        a[p]=x;
        modify(1,p,p,k);
        if(p<n) modify(1,p+1,n,-k);
        query(1);
        printf("%lld\n",ans);
    }




    return 0;
}