题解链接：http://www.cygmasot.com/index.php/2015/08/15/hdu_5381/
题意:
给定n长的序列
下面n个数给出这个序列
m个询问
下面m行给出询问的区间。
对于一个询问，输出这个区间内的任意子段的gcd 和。
思路：
因为一个数的gcd只会不变或下降，下降一次至少减半，下降至多32次，所以处理出每个数连续相同的gcd的区间。
然后暴力跑莫队。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <vector>
#include <string>
#include <time.h>
#include <math.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <set>
#include <map>
const int inf = 1e9;
const double eps = 1e-8;
const double pi = acos(-1.0);
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x < 0) { putchar('-'); x = -x; }
if (x > 9) pt(x / 10);
putchar(x % 10 + '0');
}
using namespace std;
const int N = 1e5 + 10;
typedef long long ll;
typedef pair<int, int> pii;
template <class T>
inline T gcd(T a, T b) {
if (a > b)swap(a, b);
while (a)b %= a, swap(a, b);return b;
}
vector<int>G[N];
class MST {
struct Edge {
int from, to, dis;
Edge(int _from = 0, int _to = 0, int _dis = 0) :from(_from), to(_to), dis(_dis) {}
bool operator < (const Edge &x) const { return dis < x.dis; }
}edge[N << 3];
int f[N], tot;
int find(int x) { return x == f[x] ? x : f[x] = find(f[x]); }
bool Union(int x, int y) {
x = find(x); y = find(y);
if (x == y)return false;
if (x > y)swap(x, y);
f[x] = y;
return true;
}
public:
void init(int n) {
for (int i = 0; i <= n; i++)f[i] = i;
tot = 0;
}
void add(int u, int v, int dis) {
edge[tot++] = Edge(u, v, dis);
}
ll work() {//计算最小生成树，返回花费
sort(edge, edge + tot);
ll cost = 0;
for (int i = 0; i < tot; i++)
if (Union(edge[i].from, edge[i].to)) {
cost += edge[i].dis;
G[edge[i].from].push_back(edge[i].to);
G[edge[i].to].push_back(edge[i].from);
}
return cost;
}
}mst;
struct Point {//二维平面的点
int x, y, id;
bool operator < (const Point&a) const {
return x == a.x ? y < a.y : x < a.x;
}
}p[N];
bool cmp_id(const Point&a, const Point&b) {
return a.id < b.id;
}
class BIT {//树状数组
int c[N], id[N], maxn;
int lowbit(int x) { return x&-x; }
public:
void init(int n) {
maxn = n + 10;
fill(c, c + maxn + 1, inf);
fill(id, id + maxn + 1, -1);
}
void updata(int x, int val, int _id) {
while (x) {
if (val < c[x]) { c[x] = val; id[x] = _id; }
x -= lowbit(x);
}
}
int query(int x) {
int val = inf, _id = -1;
while (x <= maxn) {
if (val > c[x]) { val = c[x]; _id = id[x]; }
x += lowbit(x);
}
return _id;
}
}tree;
inline bool cmp(int *x, int *y) { return *x < *y; }
class Manhattan_MST {
int A[N], B[N];
public:
ll work(int l, int r) {
for (int i = l; i <= r; i++)G[i].clear();
mst.init(r);
for (int dir = 1; dir <= 4; dir++) {
if (dir % 2 == 0)for (int i = l; i <= r; i++)swap(p[i].x, p[i].y);
else if (dir == 3)for (int i = l; i <= r; i++)p[i].y = -p[i].y;
sort(p + l, p + r + 1);
for (int i = l; i <= r; i++) A[i] = B[i] = p[i].y - p[i].x; //离散化
sort(B + l, B + r + 1);
int sz = unique(B + l, B + r + 1) - B;
//初始化反树状数组
tree.init(sz);
for (int i = r; i >= l; i--)
{
int pos = lower_bound(B + l, B + sz, A[i]) - B;
int id = tree.query(pos);
if (id != -1)
mst.add(p[i].id, p[id].id, abs(p[i].x - p[id].x) + abs(p[i].y - p[id].y));
tree.updata(pos, p[i].x + p[i].y, i);
}
}
for (int i = l; i <= r; i++)p[i].y = -p[i].y;
return mst.work();
}
}m_mst;
int n, m, a[N];
int l[N], r[N];
vector<pii>L[N], R[N];
ll ans[N], now;
int now_l, now_r;
ll cal_l(int point, int lim) {//point < lim
int pre = point; ll hehe = 0;
for (auto v : R[point]) {
hehe += (ll)v.second * (min(v.first, lim) - pre + 1);
pre = v.first + 1;
if (pre > lim)break;
}
return hehe;
}
ll cal_r(int point, int lim) {//lim < point
int pre = point; ll hehe = 0;
for (auto v : L[point]) {
hehe += (ll)v.second * (pre - max(v.first, lim) + 1);
pre = v.first - 1;
if (pre < lim)break;
}
return hehe;
}
void add(int x, int y) {
if (now_l > now_r)
{
now = a[x];
now_l = now_r = x;
x++;
if (x > y)return;
}
for (int i = x; i <= y; i++)
{
if (y < now_l)
now += cal_l(i, now_r);
else
now += cal_r(i, now_l);
}
if (y < now_l)now_l = x;
else now_r = y;
}
void del(int x, int y) {
for (int i = x; i <= y; i++)
{
if (y < now_r)
now -= cal_l(i, now_r);
else
now -= cal_r(i, now_l);
}
if (y < now_r)now_l = y + 1;
else now_r = x - 1;
}
void dfs(int u, int fa) {
if (u == fa)
add(l[u], r[u]);
else
{
if (l[u] < l[fa]) add(l[u], l[fa] - 1);
if (r[u] > r[fa]) add(r[fa] + 1, r[u]);
if (l[u] > l[fa]) del(l[fa], l[u] - 1);
if (r[u] < r[fa]) del(r[u] + 1, r[fa]);
}
ans[u] = now;
for (int v : G[u]) if (v != fa)dfs(v, u);
if (fa != u)
{
if (l[u] > l[fa]) add(l[fa], l[u] - 1);
if (r[u] < r[fa]) add(r[u] + 1, r[fa]);
if (l[u] < l[fa]) del(l[u], l[fa] - 1);
if (r[u] > r[fa]) del(r[fa] + 1, r[u]);
}
}
void deal_vector(vector<pii>&x) {
for (int i = 1; i < x.size(); i++)
if (x[i].first == x[i - 1].first)x[i] = x[i - 1];
x.erase(unique(x.begin(), x.end()), x.end());
}
void get_left_gcd() {
vector<pii>tmp;
for (int i = 1; i <= n; i++)
{
int gc = a[i];
for (int j = tmp.size() - 1; j >= 0; j--)
gc = tmp[j].first = gcd(tmp[j].first, gc);
tmp.push_back({ a[i], i });
deal_vector(tmp);
L[i] = tmp;
reverse(L[i].begin(), L[i].end());
for (int j = 0; j < L[i].size(); j++)swap(L[i][j].first, L[i][j].second);
}
}
void get_right_gcd() {
vector<pii>tmp;
for (int i = n; i; i--)
{
int gc = a[i];
for (int j = tmp.size() - 1; j >= 0; j--)
gc = tmp[j].first = gcd(tmp[j].first, gc);
tmp.push_back({ a[i], i });
deal_vector(tmp);
R[i] = tmp;
reverse(R[i].begin(), R[i].end());
for (int j = 0; j < R[i].size(); j++)swap(R[i][j].first, R[i][j].second);
}
}
int main() {
int T; rd(T);
while (T--) {
rd(n);
for (int i = 1; i <= n; i++)rd(a[i]);
get_left_gcd();
get_right_gcd();
rd(m);
for (int i = 1; i <= m; i++) {
rd(p[i].x); rd(p[i].y); p[i].id = i;
l[i] = p[i].x; r[i] = p[i].y;
}
m_mst.work(1, m);
now_l = 1; now_r = 0;
dfs(1, 1);
for (int i = 1; i <= m; i++)pt(ans[i]), puts("");
}
return 0;
}
/*
1
9
74 93 61 58 17 35 26 65 83
4
6 6
8 9
2 8
2 7
1
6
476 961 584 469 858 930
9
5 6
4 5
4 4
2 2
1 6
4 4
2 6
4 4
4 4
1
8
855 814 483 780 214 518 462 693
8
8 8
7 8
7 7
6 6
1 1
2 3
6 8
8 8
*/
The sum of gcd Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 393 Accepted Submission(s): 165
Problem Description
You have an array A ,the length of A is n
Let f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....aj)
Let f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....aj)
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≤109
1≤l<r≤n
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
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
