2019icpc徐州现场赛 H Yuuki and a problem (树状数组套主席树)

题意

2e5的数组，q个操作

1.将\(a[x]\)改为y

2.求下标l到r内所有的\(a[i]\)通过加法不能构成的最小的值

思路

通过二操作可以知道需要提取l到r内的值及其数量，而提取下标为l到r内的元素是一定要用主席树的

而用树状数组套上主席树即可实现修改操作

剩下需要解决的就是二操作：

首先只有有至少一个1，才能构成1

假设已经可以构成[1,x]，设当前区间内值为[1,x+1]的和为sum

那显然我们就能构成[1,sum]了，如果sum==x，那么答案就是x+1

这个过程可以直接暴力，最坏情况下当前区间里的数是{1,2,3,5,8,13...}，在第27项就到到2e5了，所以最多跑27次

那么复杂度O(\(27qlog^2n\))

这题其实比较友好，不需要离散化，也不需要优化空间，赞

比赛的时候其实也很可写，可惜

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;
typedef pair<ll,int> PIL;

const db eps = 1e-2;
const int mod = 1e9+7;
const int maxn = 2e5+100;
const int maxm = maxn*150;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int n,m;
inline int read(){
    int num;
    char ch;
    while((ch=getchar())<'0' || ch>'9');
    num=ch-'0';
    while((ch=getchar())>='0' && ch<='9'){
        num=num*10+ch-'0';
    }
    return num;
}
struct qst{
    int op;
    int x,y;
}prb[maxn];
ll a[maxn];
int totn,tot;
int rrot[maxn];
int ls[maxm],rs[maxm];
ll dat[maxm];
int root[maxn];
inline void insert(int &now, int l, int r, int x, int val){
    if(!now)now=++tot;
    dat[now]+=val;
    if(l==r)return;
    int mid = (l+r)>>1;
    if(x<=mid)insert(ls[now],l,mid,x,val);
    else insert(rs[now],mid+1,r,x,val);
}
int totL,totR;
int L[maxn],R[maxn];
inline ll ask(int l, int r, int k){
    if(r==k){
        ll ans=0;
        for(int i = 1; i <= totL; i++)ans-=dat[L[i]];
        for(int i = 1; i <= totR; i++)ans+=dat[R[i]];
        return ans;
    }
    int mid=(l+r)>>1;
    if(k<=mid){
        for(int i = 1; i <= totL; i++)L[i]=ls[L[i]];
        for(int i = 1; i <= totR; i++)R[i]=ls[R[i]];
        return ask(l,mid,k);
    }
    else{
        ll ans = 0;
        for(int i = 1; i <= totL; i++)ans-=dat[ls[L[i]]];
        for(int i = 1; i <= totR; i++)ans+=dat[ls[R[i]]];

        for(int i = 1; i <= totL; i++)L[i]=rs[L[i]];
        for(int i = 1; i <= totR; i++)R[i]=rs[R[i]];
        return ans+ask(mid+1,r,k);
    }
}
inline int lowbit(int x){return x&-x;}
int main() {
    int q;
    scanf("%d %d",&n, &q);
    for(int i = 1; i <= n; i++){
        a[i]=1ll*read();
    }
    totn=200000;
    for(int i = 1; i <= n; i++){
        int t = a[i];
        for(int j = i; j <= n; j+=lowbit(j)){
            insert(root[j],1,totn,t,a[i]);
        }
    }
    for(int i = 1; i <= q; i++){
        prb[i].op=read();prb[i].x=read();prb[i].y=read();
        if(prb[i].op==1){
            ll t = a[prb[i].x];
            int tt = prb[i].y;
            a[prb[i].x]=prb[i].y;
            for(int j = prb[i].x; j<=n; j+=lowbit(j)){
                insert(root[j],1,totn,t,-t);
                insert(root[j],1,totn,tt,tt);
            }
        }
        else{
            prb[i].x--;

            ll now = 1;
            ll sum = 0;
            while(1){
                totL=totR=0;
                for(int j = prb[i].x; j; j-=lowbit(j))L[++totL]=root[j];
                for(int j = prb[i].y; j; j-=lowbit(j))R[++totR]=root[j];

                int t = min(now,200000ll);
                ll tmp = ask(1,totn,t);
                if(tmp==sum){
                    printf("%lld\n",now);
                    break;
                }
                sum=tmp;now=sum+1;
            }
        }
    }
    return 0;
}