T1 自然数

首先预处理以一为左端点所有的 mex 值，然后插入线段树中。

考虑如何修改，左端点右移一位，相当于把那一位的数删掉了，记录下一个出现这个数的位置为 pos ，那么i 到 pos 之间所有大于 x 的 mex 都要修改为 x。

剩下的就是线段树基本操作了。

#include<bits/stdc++.h>
#define N 200010
#define int long long
using namespace std;
int n,jie,a[N];
long long ans;
int tree[N<<2],sum[N<<2],minn[N<<2],tag[N<<2],tong[N],ji,c[N],nxt[N],maxn[N<<2];
struct jj 
{int ai,id;}p[N];
inline void pushup(int x)
{   sum[x]=sum[x<<1]+sum[x<<1|1];
    minn[x]=min(minn[x<<1],minn[x<<1|1]);
    maxn[x]=max(maxn[x<<1],maxn[x<<1|1]);
}#include<bits/stdc++.h>
#define N 200010
#define int long long
using namespace std;
int n,jie,a[N];
long long ans;
int tree[N<<2],sum[N<<2],minn[N<<2],tag[N<<2],tong[N],ji,c[N],nxt[N],maxn[N<<2];
struct jj 
{int ai,id;}p[N];
inline void pushup(int x)
{   sum[x]=sum[x<<1]+sum[x<<1|1];
    minn[x]=min(minn[x<<1],minn[x<<1|1]);
    maxn[x]=max(maxn[x<<1],maxn[x<<1|1]);
}
inline void pushdown(int x,int l,int r)
{   minn[x<<1]=minn[x<<1|1]=maxn[x<<1]=maxn[x<<1|1]=tag[x<<1]=tag[x<<1|1]=tag[x];
    int mid=(l+r)>>1;sum[x<<1]=tag[x]*(mid-l+1);sum[x<<1|1]=tag[x]*(r-mid);tag[x]=-1;
}
inline void build(int x,int l,int r)
{   if(l==r){sum[x]=minn[x]=maxn[x]=c[l];return;}
    int mid=(l+r)>>1;
    build(x<<1,l,mid);build(x<<1|1,mid+1,r);
    pushup(x);
}
inline int query(int x,int l,int r,int L,int R)
{   if(l>=L and r<=R)return sum[x];
    int mid=(l+r)>>1,res=0;if(tag[x]!=-1)pushdown(x,l,r);
    if(mid<R)res+=query(x<<1|1,mid+1,r,L,R);
    if(mid>=L)res+=query(x<<1,l,mid,L,R);
    return res;
}
inline void update(int x,int l,int r,int L,int R,int val)
{   if(L>R)return;
    if(l>=L and r<=R and minn[x]>=val){maxn[x]=minn[x]=val;sum[x]=val*(r-l+1);tag[x]=val;return;}
    int mid=(l+r)>>1;if(tag[x]!=-1)pushdown(x,l,r);
    if(mid<R and maxn[x<<1|1]>val)update(x<<1|1,mid+1,r,L,R,val);
    if(mid>=L and maxn[x<<1]>val)update(x<<1,l,mid,L,R,val);
    pushup(x);
}
signed main()
{   freopen("mex.in","r",stdin);
    freopen("mex.out","w",stdout);
    scanf("%lld",&n);
    for(int i=1;i<=n;++i)scanf("%lld",&a[i]),p[i]=(jj){a[i],i};
    int zhi=0;
    for(int i=1;i<=n;++i)
    {   if(a[i]<=n)tong[a[i]]=1;
        while(tong[zhi])++zhi;
        c[i]=zhi;
    }
    memset(tong,0,sizeof(tong));memset(tag,-1,sizeof(tag));
    for(int i=n;i;--i)if(a[i]<=n){nxt[i]=tong[a[i]];tong[a[i]]=i;} 
    build(1,1,n);
    for(int i=1;i<=n;++i)
    {   ans+=query(1,1,n,i,n);
        if(a[i]<=n)
        {   int pos=nxt[i];
            if(!pos)pos=n+1;
            update(1,1,n,i,pos-1,a[i]);  
        }
    }
    printf("%lld\n",ans);
}
    build(x<<1,l,mid);build(x<<1|1,mid+1,r);
    pushup(x);
}
inline int query(int x,int l,int r,int L,int R)
{   if(l>=L and r<=R)return sum[x];
    int mid=(l+r)>>1,res=0;if(tag[x]!=-1)pushdown(x,l,r);
    if(mid<R)res+=query(x<<1|1,mid+1,r,L,R);
    if(mid>=L)res+=query(x<<1,l,mid,L,R);
    return res;
}
inline void update(int x,int l,int r,int L,int R,int val)
{   if(L>R)return;
    if(l>=L and r<=R and minn[x]>=val){maxn[x]=minn[x]=val;sum[x]=val*(r-l+1);tag[x]=val;return;}
    int mid=(l+r)>>1;if(tag[x]!=-1)pushdown(x,l,r);
    if(mid<R and maxn[x<<1|1]>val)update(x<<1|1,mid+1,r,L,R,val);
    if(mid>=L and maxn[x<<1]>val)update(x<<1,l,mid,L,R,val);
    pushup(x);
}
signed main()
{   freopen("mex.in","r",stdin);
    freopen("mex.out","w",stdout);
    scanf("%lld",&n);
    for(int i=1;i<=n;++i)scanf("%lld",&a[i]),p[i]=(jj){a[i],i};
    int zhi=0;
    for(int i=1;i<=n;++i)
    {   if(a[i]<=n)tong[a[i]]=1;
        while(tong[zhi])++zhi;
        c[i]=zhi;
    }
    memset(tong,0,sizeof(tong));memset(tag,-1,sizeof(tag));
    for(int i=n;i;--i)if(a[i]<=n){nxt[i]=tong[a[i]];tong[a[i]]=i;} 
    build(1,1,n);
    for(int i=1;i<=n;++i)
    {   ans+=query(1,1,n,i,n);
        if(a[i]<=n)
        {   int pos=nxt[i];
            if(!pos)pos=n+1;
            update(1,1,n,i,pos-1,a[i]);  
        }
    }
    printf("%lld\n",ans);
}

T2 钱仓

贪心，一定存在中转点，一定存在一个分界点满足只向一遍运输，这个分界点就是最大子段和的起点。

#include<bits/stdc++.h>
#define int long long
#define N 200005
using namespace std;
int n,c[N],sum[N],nxt[N],maxn,id,tmp,t[N],ans,old[N];
signed main()
{   freopen("barn.in","r",stdin);
    freopen("barn.out","w",stdout);
    scanf("%lld",&n);
    for(int i=1;i<=n;++i)scanf("%lld",&c[i]);
    for(int i=1;i<=n;++i)
    {   tmp+=c[i];
        if(tmp<i-id)id=i,tmp=0;
    }
    ++id;
    for(int i=id;i<=n;++i)t[i-id+1]=c[i];for(int i=1;i<id;++i)t[i+n-id+1]=c[i];
    int zhi1=n;
    for(int i=n;i;--i)
    {   if(zhi1>i)zhi1=i;
        while(!t[zhi1])--zhi1;
        if(!t[i])ans+=(i-zhi1)*(i-zhi1),--t[zhi1];
    }
    printf("%lld\n",ans);
}

T3

标签：七十四,zhi1,val,NOIP,int,long,tag,模拟,define
来源： https://www.cnblogs.com/zhaoxubing/p/15396019.html