bzoj 5110 Yazid的新生舞会

题目大意：

一个数列，求有多少个区间$[l,r]$满足该区间的众数出现次数大于$\lceil \frac{r-l}{2} \rceil$

思路：

对于一个区间满足条件的众数明显是唯一的 所以设该数的前缀和数组为$S$

则一个区间$(l,r]$满足条件满足$2 \times (S_r-S_l)>(r-l)$ 移项后得到$2 \times S_r-r>2 \times S_l-l$

则对于$2 \times S_i-i$建立函数发现该函数由斜率为$\pm 1$组成

所有函数的关键点总数为$n$ 对于$y$轴建立权值线段树 发现中间连续的一段的贡献为一次函数

写一波一次函数即可

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<vector>
 #include<queue>
 #include<map>
 #include<set>
 #define rep(i,s,t) for(register int i=(s),i__end=(t);i<=i__end;++i)
 #define dwn(i,s,t) for(register int i=(s),i__end=(t);i>=i__end;++i)
 #define ren(x) for(register int i=fst[x];i;i=nxt[i])
 #define pb(a,x) vec[a].push_back(x);
 #define ll long long
 #define inf 2139062143
 #define MAXN 1001000
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(!isdigit(ch)) {if(ch=='-') f=-;ch=getchar();}
     while(isdigit(ch)) {x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,N,tp,g[MAXN],hsh[MAXN],m,tagk[MAXN<<],tagb[MAXN<<];
 ll sum[MAXN<<],ans;
 vector <int> vec[MAXN];
 inline ll calc(ll k,ll b,int x,int y) {return ((ll)((ll)k*y+k*x+*b)*((ll)y-x+))/;}
 void pshd(int k,int l,int r,int mid)
 {
     sum[k<<]+=calc(tagk[k],tagb[k],l,mid);
     sum[k<<|]+=calc(tagk[k],tagb[k],mid+,r);
     tagk[k<<]+=tagk[k],tagk[k<<|]+=tagk[k],tagb[k<<]+=tagb[k],tagb[k<<|]+=tagb[k];
     tagk[k]=tagb[k]=0LL;
 }
 void mdf(int k,int l,int r,int a,int b,ll d,ll t)
 {
     if(l==a&&r==b) {tagb[k]+=t,tagk[k]+=d,sum[k]+=calc(d,t,l,r);return ;}
     int mid=l+r>>;if(tagb[k]!=||tagk[k]!=) pshd(k,l,r,mid);
     if(b<=mid) mdf(k<<,l,mid,a,b,d,t);else if(a>mid) mdf(k<<|,mid+,r,a,b,d,t);
     else {mdf(k<<,l,mid,a,mid,d,t);mdf(k<<|,mid+,r,mid+,b,d,t);}
     sum[k]=sum[k<<]+sum[k<<|];
 }
 ll query(int k,int l,int r,int a,int b)
 {
     if(l==a&&r==b) return sum[k];
     int mid=l+r>>;if(tagb[k]!=||tagk[k]!=) pshd(k,l,r,mid);
     if(b<=mid) return query(k<<,l,mid,a,b);
     else if(a>mid) return query(k<<|,mid+,r,a,b);
     else return query(k<<,l,mid,a,mid)+query(k<<|,mid+,r,mid+,b);
 }
 void work(int x)
 {
     int t,las=,pos;rep(i,,vec[x].size()-)
     {
         t=vec[x][i],pos=las-(t-vec[x][i-])+;
         ans+=query(,,N,pos+n,las+n);
         mdf(,,N,pos+n+,las+n+,,-pos-n);
         if(las+n+<=N) mdf(,,N,las+n+,N,,las-pos+);
         las=pos+;
     }
     las=;rep(i,,vec[x].size()-)
     {
         t=vec[x][i],pos=las-(t-vec[x][i-])+;
         mdf(,,N,pos+n+,las+n+,-,pos+n);
         if(las+n+<=N) mdf(,,N,las+n+,N,,pos-las-);
         las=pos+;
     }
 }
 int main()
 {
     n=read(),N=(n<<)+,tp=read();rep(i,,n) {g[i]=read();if(!hsh[g[i]]) hsh[g[i]]=++m;}
     rep(i,,m) pb(i,);rep(i,,n) pb(hsh[g[i]],i);rep(i,,m) pb(i,n+);
     rep(i,,m) work(i);printf("%lld\n",ans);
 }