http://acm.hdu.edu.cn/showproblem.php?pid=5726

求不修改区间gcd可以用线段树或者倍增。

求l-n的我们注意观察gcd(a​l​​,a​l+1​​,...,a​r​​),当l固定不动的时候,r=l...n时,我们可以容易的发现,随着r的増大,gcd(a​l​​,a​l+1​​,...,a​r​​)是递减的,同时gcd(a​l​​,a​l+1​​,...,a​r​​)最多 有log 1000,000,000个不同的值,因为a​l​​最多也就有log 1000,000,000个质因数。

然后我们用链表记录所有gcd改变的点,这些点将l...n这一段分成若干个相同gcd的区间。由l...n的gcd关系可以推出l-1...n的gcd关系。相邻区间gcd相同时将两个区间合并。用map统计gcd是x的区间有多少个。

 #include<cstdio>
#include<cstring>
#include<map>
#define N 100010
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
int te,l[N*],r[N*],a[N*],f[N],x[N];
long long j,n,nex[N];
int t,m,ans,ll,rr,num,i,_;
map<int,long long> sc; int gcd(int aa,int bb)
{
int tt,a=aa,b=bb;
while (a%b!=)
{
tt=a;
a=b;
b=tt%b;
}
return b;
}
void build(int s,int ll,int rr)
{
l[s]=ll;r[s]=rr;
if (ll==rr)
{
scanf("%d",&x[++num]);
a[s]=x[num];
}
else
{
build(s*,ll,(ll+rr)/);
build(s*+,(ll+rr)/+,rr);
a[s]=gcd(a[s*],a[s*+]);
}
}
void sea(int s)
{
if (l[s]>rr || r[s]<ll) return;
if (ll<=l[s] && rr>=r[s])
if (ans==-)
ans=a[s];
else
ans=gcd(ans,a[s]);
else
{
sea(s*);
sea(s*+);
}
}
int main()
{
scanf("%d",&_);
while (_--)
{
printf("Case #%d:\n",++te);
num=;
mem(f);sc.clear();mem(nex);mem(l);mem(r);mem(a);
scanf("%d",&n);
build(,,n);
f[n]=x[n];sc[x[n]]++;
for (i=n-;i>=;i--)
{
j=i+;
while (j!=)
{
f[j]=gcd(f[j],x[i]);
j=nex[j];
}
f[i]=x[i];
nex[i]=i+;
j=i;
while (nex[j]!=)
{
if (f[j]==f[nex[j]])
nex[j]=nex[nex[j]];
else
j=nex[j];
}
j=i;
while (j!=)
{
if (nex[j]==)
sc[f[j]]+=n-j+;
else
sc[f[j]]+=nex[j]-j;
j=nex[j];
} }
scanf("%d",&m);
for (t=;t<=m;t++)
{
scanf("%d%d",&ll,&rr);
ans=-;
sea();
printf("%d %I64d\n",ans,sc[ans]);
}
}
return ;
}
Run ID Submit Time Judge Status Pro.ID Exe.Time Exe.Memory Code Len. Language Author
17643445 2016-07-20 14:35:10 Accepted 5726 1950MS 14112K 1557B G++ lbz007

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