[分治] BZOJ3745: [Coci2015]Norma

显然是分治套路题。分段求，搞一堆的前缀和就行了。

#include<cstdio>#include<algorithm>using namespace std;typedef long long LL;const int maxn=500005,MOD=1e+9;int n,a[maxn];LL _min[maxn],_max[maxn],s1[maxn],s2[maxn],s12[maxn],s1i[maxn],s2i[maxn],s12i[maxn];LL ans;void Solve(int L,int R){    if(L==R){ (ans+=(LL)a[L]*a[L]%MOD)%=MOD; return; }    int mid=(L+R)>>1;    Solve(L,mid); Solve(mid+1,R);    _min[mid]=1e+9; _max[mid]=0; s1[mid]=s2[mid]=s1i[mid]=s2i[mid]=s12[mid]=s12i[mid]=0;    for(int i=mid+1;i<=R;i++){        _min[i]=min(_min[i-1],(LL)a[i]); _max[i]=max(_max[i-1],(LL)a[i]);        s1[i]=(s1[i-1]+_min[i])%MOD; s2[i]=(s2[i-1]+_max[i])%MOD; s12[i]=(s12[i-1]+_max[i]*_min[i])%MOD;        s1i[i]=(s1i[i-1]+_min[i]*(i-mid))%MOD; s2i[i]=(s2i[i-1]+_max[i]*(i-mid))%MOD; s12i[i]=(s12i[i-1]+_max[i]*_min[i]%MOD*(i-mid)%MOD)%MOD;    }    for(int i=mid,maxL=0,minL=1e+9,p1=mid,p2=mid;i>=L;i--){        minL=min(minL,a[i]); while(p1<R&&_min[p1+1]>minL) p1++;        maxL=max(maxL,a[i]); while(p2<R&&_max[p2+1]<maxL) p2++;        int k=min(p1,p2); (ans+=((LL)(mid+1-i+1+k-i+1)*(k-mid)/2)%MOD*minL%MOD*maxL%MOD)%=MOD;        if(p1<p2) (ans+=(s1i[p2]-s1i[p1]+(s1[p2]-s1[p1])*(mid-i+1))%MOD*maxL)%=MOD;              else (ans+=(s2i[p1]-s2i[p2]+(s2[p1]-s2[p2])*(mid-i+1))%MOD*minL)%=MOD;        k=max(p1,p2); (ans+=(s12i[R]-s12i[k])+(s12[R]-s12[k])*(mid-i+1))%=MOD;    }}int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++) scanf("%d",&a[i]);    Solve(1,n);    printf("%lld\n",(ans+MOD)%MOD);    return 0;}