HDU 5381 The sum of gcd 莫队算法

题目：http://acm.hdu.edu.cn/showproblem.php?pid=5381

题意：给定一个数组，每次询问给出l和r，求出 

思路：看的大牛的代码，心累

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#define debug() puts("here")
using namespace std;

typedef long long ll;
const int N = 10010;
struct node
{
    int l, r, id;
} g[N];
int n, m, unit;
ll res[N], tmp;
int a[N], b[N], c[N];
struct vv
{
    int id, g;
};
vector<vv> vec1[N], vec2[N];
int gcd(int a, int b)
{
    return !b ? a : gcd(b, a % b);
}
bool cmp(node a, node b)
{
    return a.l/unit != b.l/unit ? a.l/unit < b.l/unit : a.r < b.r;
}
void work()
{
    for(int i = 1; i <= n; i++) vec1[i].clear(), vec2[i].clear();
    for(int i = 1; i <= n; i++)
    {
        if(i == 1) vec1[i].push_back(vv{i, a[i]});
        else
        {
            int curg = a[i], id = i;
            for(auto &it : vec1[i-1])
            {
                int g = gcd(curg, it.g);
                if(g != curg) vec1[i].push_back(vv{id, curg});
                curg = g, id = it.id;
            }
            vec1[i].push_back(vv{id, curg});
        }
    }
    for(int i = n; i >= 1; i--)
    {
        if(i == n) vec2[i].push_back(vv{i, a[i]});
        else
        {
            int curg = a[i], id = i;
            for(auto &it : vec2[i+1])
            {
                int g = gcd(curg, it.g);
                if(g != curg) vec2[i].push_back(vv{id, curg});
                curg = g, id = it.id;
            }
            vec2[i].push_back(vv{id, curg});
        }
    }
}
ll calr(int l, int r)
{
    ll sum = 0;
    int tr = r;
    for(auto &it : vec1[r])
        if(it.id >= l)
        {
            sum += (ll)(tr - it.id + 1) * it.g;
            tr = it.id - 1;
        }
        else
        {
            sum += (ll)(tr - l + 1) * it.g;
            break;
        }
    return sum;
}
ll call(int l, int r)
{
    ll sum = 0;
    int tl = l;
    for(auto &it : vec2[l])
        if(it.id <= r)
        {
            sum += (ll)(it.id - tl + 1) * it.g;
            tl = it.id + 1;
        }
        else
        {
            sum += (ll)(r - tl + 1) * it.g;
            break;
        }
    return sum;
}
void solve()
{
    unit = (int)sqrt(1.0 * n);
    sort(g+1, g+1+m, cmp);
    int l = 1, r = 0;
    tmp = 0;
    for(int i = 1; i <= m; i++)
    {
        while(r < g[i].r) tmp += calr(l, ++r);
        while(r > g[i].r) tmp -= calr(l, r--);
        while(l < g[i].l) tmp -= call(l++, r);
        while(l > g[i].l) tmp += call(--l, r);
        res[g[i].id] = tmp;
    }
    for(int i = 1; i <= m; i++) printf("%lld\n", res[i]);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        scanf("%d", &m);
        for(int i = 1; i <= m; i++) scanf("%d%d", &g[i].l, &g[i].r), g[i].id = i;
        work();
        solve();
    }
    return 0;
}