【NOIP2015模拟11.5】JZOJ8月5日提高组T2 Lucas的数列
【NOIP2015模拟11.5】JZOJ8月5日提高组T2 Lucas的数列
题目
PS:\(n*n*T*T<=10^{18}\)而不是\(10^1*8\)
题解
题意:
给出\(n\)个元素的复杂度和位置
然后每次询问一个区间
每次询问给出\(x,y,z\)
求\(x\)到\(y\)中复杂度小于等于\(z\)的元素的位置按照一定计算过程后的答案(具体见题面)
分析:
看到题目给出的式子十分的繁琐
我们来化简一下
\[K=(\sum_{i=1}^m(x_i-p)^2)*m=(\sum_{i=1}^m((x_i)^2-2x_ip+p^2))*m=(\sum_{i=1}^m(x_i)^2-\sum_{i=1}^m2x_ip+\sum_{i=1}^mp^2)*m
\]
\[=(\sum_{i=1}^m(x_i)^2-\sum_{i=1}^m2x_i\dfrac{\sum_{i=1}^mx_i}{m}+m(\dfrac{\sum_{i=1}^mx_i}{m})^2)*m=m\sum_{i=1}^m(x_i)^2-2({\sum_{i=1}^mx_i})^2+({\sum_{i=1}^mx_i})^2=m\sum_{i=1}^m(x_i)^2-({\sum_{i=1}^mx_i})^2
\]
所以说,\(K\)其实一直都是个整数:\(m\sum_{i=1}^m(x_i)^2-({\sum_{i=1}^mx_i})^2\)
再看,这题并不要求在线
所以可以离线
按照\(w\)和\(z\)为第一关键字排序
然后维护一个\(j\)使得\(1\)~\(j\)内的\(w\)都小于当前的\(z\)
由于\(z\)是单调递增的,所以\(j\)不用清零
然后构造一棵线段树(树状数组)就可以了
Code
#include<cstdio>
#include<algorithm>
using namespace std;
struct node1
{
long long p,w,id;
}a[400005];
struct node2
{
long long x,y,z,lixian;
}c[400005];
struct node3
{
long long sum1,sum2,num;
}tree[1600005];
bool cmp1(node1 x,node1 y)
{
return x.w<y.w;
}
bool cmp2(node2 x,node2 y)
{
return x.z<y.z;
}
long long n,m,i,j;
long long s1,s2,s3,ans[400005];
void build(long long now,long long l,long long r,long long pos,long long val)
{
if (l==r)
{
if (l==pos)
{
tree[now].num=1;
tree[now].sum1=val*val;
tree[now].sum2=val;
}
return;
}
if (l>pos||r<pos) return;
long long mid=(l+r)>>1;
build(now<<1,l,mid,pos,val);
build(now<<1|1,mid+1,r,pos,val);
tree[now].num=tree[now<<1].num+tree[now<<1|1].num;
tree[now].sum1=tree[now<<1].sum1+tree[now<<1|1].sum1;
tree[now].sum2=tree[now<<1].sum2+tree[now<<1|1].sum2;
}
void query(long long now,long long l,long long r,long long p,long long q)
{
if (tree[now].num==0) return;
if (l>q||r<p) return;
if (l>=p&&r<=q)
{
s1+=tree[now].num;
s2+=tree[now].sum1;
s3+=tree[now].sum2;
return;
}
long long mid=(l+r)>>1;
query(now<<1,l,mid,p,q);
query(now<<1|1,mid+1,r,p,q);
}
int main()
{
freopen("sequence.in","r",stdin);
freopen("sequence.out","w",stdout);
scanf("%lld%lld",&n,&m);
for (i=1;i<=n;i++)
{
scanf("%lld%lld",&a[i].w,&a[i].p);
a[i].id=i;
}
for (i=1;i<=m;i++)
{
scanf("%lld%lld%lld",&c[i].x,&c[i].y,&c[i].z);
c[i].lixian=i;
}
sort(a+1,a+n+1,cmp1);
sort(c+1,c+m+1,cmp2);
j=1;
for (i=1;i<=m;i++)
{
while (a[j].w<=c[i].z&&j<=n)
{
build(1,1,n,a[j].id,a[j].p);
j++;
}
s1=s2=s3=0;
query(1,1,n,c[i].x,c[i].y);
if (s1==0) ans[c[i].lixian]=-1;
else ans[c[i].lixian]=s1*s2-s3*s3;
}
for (i=1;i<=m;i++)
{
if (ans[i]==-1) printf("empty\n");
else printf("%lld\n",ans[i]);
}
fclose(stdin);
fclose(stdout);
return 0;
}