[分治] BZOJ 3745 [Coci2015]Norma

最新推荐文章于 2019-10-06 20:47:44 发布

原创最新推荐文章于 2019-10-06 20:47:44 发布 · 709 阅读

0 ·

CC 4.0 BY-SA版权

分治专栏收录该内容

16 篇文章

订阅专栏

本文介绍了一种使用分治策略解决特定问题的方法。通过将问题分解为较小的子问题，并利用前缀和等技术来高效地求解这些子问题，最终得到整个问题的解答。文章详细展示了算法的具体实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

考虑分治

左端点在左边右端点在右边

枚举左端点分情况讨论最大最小分别在哪边然后前缀和处理

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef long long ll;

inline char nc(){
    static char buf[100000],*p1=buf,*p2=buf;
    if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
    return *p1++;
}

inline void read(ll &x){
    char c=nc(),b=1;
    for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
    for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const ll P=(ll)1e9;
const int N=500005;

ll n,a[N];
ll sum1[N],sum2[N],sum3[N],sum4[N],sum5[N],sum6[N];

inline ll Sum(ll s,ll t){
    return ((s+t)*(t-s+1)/2)%P;
}

ll ans;

inline void Solve(int l,int r){
    if (l==r){
        ans+=a[l]*a[l]%P; ans=(ans%P+P)%P; return;
    }
    int mid=(l+r)>>1;
    ll mx=-1LL<<30,mn=1LL<<30; sum1[mid]=sum2[mid]=sum3[mid]=sum4[mid]=sum5[mid]=sum6[mid]=0;
    for (int i=mid+1;i<=r;i++)
    {
        mx=max(mx,a[i]); mn=min(mn,a[i]);
        sum1[i]=mx*mn%P;        (sum1[i]+=sum1[i-1])%=P;
        sum2[i]=mx*mn%P*i%P;    (sum2[i]+=sum2[i-1])%=P;
        sum3[i]=mn;             (sum3[i]+=sum3[i-1])%=P;
        sum4[i]=mn*i%P;         (sum4[i]+=sum4[i-1])%=P;
        sum5[i]=mx;             (sum5[i]+=sum5[i-1])%=P;
        sum6[i]=mx*i%P;         (sum6[i]+=sum6[i-1])%=P;
    }
    int j=mid,k=mid; ll u=a[mid],v=a[mid],s,t;
    for (int i=mid;i>=l;i--,u=min(u,a[i]),v=max(v,a[i])){
        while (j+1<=r && a[j+1]>=u) j++;
        while (k+1<=r && a[k+1]<=v) k++;
        s=mid+1,t=min(j,k);
        if (s<=t)
        {
            ans+=u*v%P*Sum(s-i+1,t-i+1)%P; ans=(ans%P+P)%P;
        }
        s=max(j,k)+1,t=r;
        if (s<=t)
        {
            ans+=sum2[t]-sum2[s-1]; ans=(ans%P+P)%P;
            ans+=(ll)(1-i)*(sum1[t]-sum1[s-1])%P; ans=(ans%P+P)%P;
        }
        s=min(j,k)+1,t=max(j,k);
        if (s<=t)
        {
            if (j<=k){
                ans+=v*(sum4[t]-sum4[s-1])%P; ans=(ans%P+P)%P;
                ans+=v*(1-i)%P*(sum3[t]-sum3[s-1])%P; ans=(ans%P+P)%P;
            }else{
                ans+=u*(sum6[t]-sum6[s-1])%P; ans=(ans%P+P)%P;
                ans+=u*(1-i)%P*(sum5[t]-sum5[s-1])%P; ans=(ans%P+P)%P;
            }
        }
    }
    Solve(l,mid); Solve(mid+1,r);
}

int main()
{
    freopen("t.in","r",stdin);
    freopen("t.out","w",stdout);
    read(n);
    for (int i=1;i<=n;i++) read(a[i]);
    Solve(1,n);
    printf("%lld\n",ans);
    return 0;
}