The sum of gcd
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 784 Accepted Submission(s): 335
Problem Description
You have an array A,the
length of A
is n
Let f(l,r)=∑r
i=l
∑
r
j=i
gcd(a
i
,a
i+1
....a
j
)
Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
First line has one integers n
Second line has n
integers Ai
Third line has one integers Q,the
number of questions
Next there are Q lines,each line has two integers l,r
1≤T≤3
1≤n,Q≤104
1≤ai
≤10
9
1≤l<r≤n
Output
For each question,you need to print
f(l,r)
Sample Input
2 5 1 2 3 4 5 3 1 3 2 3 1 4 4 4 2 6 9 3 1 3 2 4 2 3
Sample Output
9 6 16 18 23 10
Author
SXYZ
Source
2015 Multi-University Training Contest 8
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预处理一段的gcd
我们发现一段的gcd是一样的
1 2 3 6 6 6 12
1 1 3 6 6 6 12 //从12开始向左的gcd
显然最多有logN段，接下来用莫队算法+杨氏转移
O(nlogn+nsqrt(n)logn)=O(n^1.5*logn)
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (10000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,a[MAXN],Q;
struct seg {
int l,r,i;
friend bool operator<(seg a,seg b){ return (int)((a.l)/sqrt(n))^(int)((b.l)/sqrt(n))?(int)((a.l)/sqrt(n))<(int)((b.l)/sqrt(n)):a.r<b.r;
}
}comm[MAXN];
ll ans[MAXN];
ll gcd(ll a,ll b){if (b==0) return a; return gcd(b,a%b);}
int h[MAXN][100]={0};
ll val[MAXN][100]={0};
void init(int h[][100],ll val[][100])
{
MEM(val) MEM(h)
For(i,n) h[i][0]=0;
h[1][0]=1; h[1][1]=1; val[1][1]=a[1];
Fork(i,2,n) {
int &k=h[i][0];
val[i][++k]=a[i];h[i][k]=i;
For(j,h[i-1][0]) {
ll p=gcd(val[i][k],val[i-1][j]);
if (p!=val[i][k])
val[i][++k]=p;
h[i][k]=h[i-1][j];
}
}
}
int h2[MAXN][100]={0};
ll val2[MAXN][100]={0};
void init2(int h[][100],ll val[][100])
{
MEM(val) MEM(h)
For(i,n) h[i][0]=0;
h[n][0]=1; h[n][1]=n; val[n][1]=a[n];
ForD(i,n-1) {
int &k=h[i][0];
val[i][++k]=a[i];h[i][k]=i;
For(j,h[i+1][0]) {
ll p=gcd(val[i][k],val[i+1][j]);
if (p!=val[i][k])
val[i][++k]=p;
h[i][k]=h[i+1][j];
}
}
}
ll modify(){return 0;
}
ll modify(int l,int r,int f)
{
ll ret=0;
if (f==0) { //left
int fro=l;
For(j,h2[l][0]) {
int last=min(h2[l][j],r);
if (fro<=last) ret+=val2[l][j]*(last-fro+1);
fro=last+1;
if (fro>r) break;
}
} else {
int last=r;
For(j,h[r][0]) {
int fro=max(h[r][j],l);
if (fro<=last) ret+=val[r][j]*(last-fro+1);
last=fro-1;
if (fro<l) break;
}
}
return ret;
}
int main()
{
// freopen("B.in","r",stdin);
int T; cin>>T;
while(T--) {
cin>>n;
For(i,n) scanf("%d",&a[i]);
init(h,val);init2(h2,val2);
cin>>Q;
MEM(ans)
For(i,Q)
{
scanf("%d%d",&comm[i].l,&comm[i].r),comm[i].i=i;
}
sort(comm+1,comm+1+Q);
int nowl=1,nowr=1;
ll nowans=a[1];
For(i,Q)
{
while (nowl<comm[i].l) nowans-=modify(nowl,nowr,0),nowl++;
while (nowl>comm[i].l) nowans+=modify(nowl-1,nowr,0),nowl--;
while (comm[i].r<nowr) nowans-=modify(nowl,nowr,1),nowr--;
while (comm[i].r>nowr) nowans+=modify(nowl,nowr+1,1),nowr++;
ans[comm[i].i]=nowans;
}
For(i,Q) printf("%I64d\n",ans[i]);
}
return 0;
}
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