【分治计数】BZOJ3745 [Coci2015]Norma

题面在这里

分治计数的经典题目……

对mid右边的每个位置维护以下量：

i−mid的前缀和

前缀max和min

前缀max和min的前缀和mx和mn

mx⋅mn的前缀和

mx⋅mn(i−mid)的前缀和

mx(i−mid)和mn(i−mid)的前缀和

然后JXB讨论一下就好了

示例程序：

#include<cstdio>#include<algorithm>using namespace std;typedef long long LL;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int red(){    int res=0,f=1;char ch=nc();    while (ch<'0'||'9'<ch) {if (ch=='-') f=-f;ch=nc();}    while ('0'<=ch&&ch<='9') res=res*10+ch-48,ch=nc();    return res*f;}const int maxn=500005,INF=0x3f3f3f3f,tt=1000000000;int n,a[maxn];LL ans,s[maxn],smx[maxn],smn[maxn],ss[maxn],ssl[maxn],smxl[maxn],smnl[maxn],mx[maxn],mn[maxn];void devide(int L,int R){    if (L==R) {(ans+=(LL)a[L]*a[R]%tt)%=tt;return;}    int mid=L+R>>1;    devide(L,mid);devide(mid+1,R);    s[mid]=smx[mid]=smn[mid]=mx[mid]=ss[mid]=ssl[mid]=smxl[mid]=smnl[mid]=0; mn[mid]=INF;    for (int i=mid+1;i<=R;i++){        mx[i]=max(mx[i-1],(LL)a[i]); mn[i]=min(mn[i-1],(LL)a[i]);        smx[i]=(smx[i-1]+mx[i])%tt; smn[i]=(smn[i-1]+mn[i])%tt;        s[i]=(s[i-1]+i-mid)%tt; ss[i]=(ss[i-1]+mx[i]*mn[i])%tt;        ssl[i]=(ssl[i-1]+mx[i]*mn[i]%tt*(i-mid))%tt;        smxl[i]=(smxl[i-1]+mx[i]*(i-mid))%tt; smnl[i]=(smnl[i-1]+mn[i]*(i-mid))%tt;    }    LL Max=0,Min=INF;    for (int i=mid,jx=mid,jm=mid;i>=L;i--){        Max=max(Max,(LL)a[i]);Min=min(Min,(LL)a[i]);        while (jx<R&&mx[jx+1]<Max) jx++;        while (jm<R&&mn[jm+1]>Min) jm++;        int x=min(jx,jm),y=max(jx,jm);LL ii=mid-i+1;        (ans+=(s[x]+ii*(x-mid))*Max%tt*Min)%=tt;        if (jx<=jm)         (ans+=(smxl[y]-smxl[x] + ii*(smx[y]-smx[x]))%tt*Min)%=tt;        else         (ans+=(smnl[y]-smnl[x] + ii*(smn[y]-smn[x]))%tt*Max)%=tt;        (ans+=ssl[R]-ssl[y] + ii*(ss[R]-ss[y]))%=tt;    }}int main(){    n=red();    for (int i=1;i<=n;i++) a[i]=red();    devide(1,n);    printf("%lld",(ans+tt)%tt);    return 0;}