2015 Multi-University Training Contest 8 The sum of gcd

The sum of gcd

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1156    Accepted Submission(s): 508


[align=left]Problem Description[/align]
You have an array A,the
length of A
is n

Let f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....aj)

 

[align=left]Input[/align]
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≤109
1≤l<r≤n
 

[align=left]Output[/align]
For each question,you need to print
f(l,r)
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

9
6
16
18
23
10

【题意】如题所示。

【解题方法】gcd预处理套莫队算法。大牛博客http://blog.csdn.net/u014800748/article/details/47680899

【AC 代码】

//
//Created by just_sort 2016/9/12 16:50
//Copyright (c) 2016 just_sort.All Rights Reserved
//

#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e4+7;
int n,m,block;
int a[maxn];
LL ans[maxn];
struct Q{
int l,r,id;
bool operator<(const Q &rhs) const{
if(l/block==rhs.l/block) return r<rhs.r;
return l/block<rhs.l/block;
}
}q[maxn];
struct node{int id;int g;};
vector <node> vl[maxn],vr[maxn];
void init(){
for(int i=1; i<=n; i++){
if(i==1) vl[i].push_back(node{i,a[i]});
else{
int curg=a[i];
int L=i;
for(auto &it:vl[i-1]){
int g = __gcd(it.g,curg);
if(g!=curg) vl[i].push_back(node{L,curg});
curg=g,L=it.id;
}
vl[i].push_back(node{L,curg});
}
}
for(int i=n; i>=1; i--){
if(i==n) vr[i].push_back(node{i,a[i]});
else{
int curg=a[i];
int R=i;
for(auto &it:vr[i+1]){
int g = __gcd(it.g,curg);
if(g!=curg) vr[i].push_back(node{R,curg});
curg=g,R=it.id;
}
vr[i].push_back(node{R,curg});
}
}
}
LL cal(int op,int L,int R)
{
LL ret = 0;
if(op==0){
int tr = R;
for(auto &it : vl[R]){
if(it.id>=L){
ret += 1LL*(tr-it.id+1)*it.g;
tr = it.id-1;
}else{
ret += 1LL*(tr-L+1)*it.g;
break;
}
}
}else{
int tl = L;
for(auto &it : vr[L]){
if(it.id<=R){
ret += 1LL*(it.id-tl+1)*it.g;
tl = it.id+1;
}else{
ret += 1LL*(R-tl+1)*it.g;
break;
}
}
}
return ret;
}

int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1; i<=n; i++) vl[i].clear(),vr[i].clear();
for(int i=1; i<=n; i++) scanf("%d",&a[i]);
init();
scanf("%d",&m);
for(int i=0; i<m; i++){
scanf("%d%d",&q[i].l,&q[i].r);
q[i].id=i;
}
block = (sqrt(n));
sort(q,q+m);
int L=1,R=0;
LL ret = 0;
for(int i=0; i<m; i++){
while(R<q[i].r){
R++;
ret += cal(0,L,R);
}
while(R>q[i].r){
ret -= cal(0,L,R);
R--;
}
while(L>q[i].l){
L--;
ret += cal(1,L,R);
}
while(L<q[i].l){
ret -= cal(1,L,R);
L++;
}
ans[q[i].id] = ret;
}
for(int i=0; i<m; i++){
printf("%I64d\n",ans[i]);
}
}
return 0;
}