洛谷 P2522 [HAOI2011]Problem b

题意：

求$sum_{x=a}^{b}sum_{y=c}^{d}[gcd(x, y) = k]$

思路：

   $ sum_{x=a}^{b}sum_{y=c}^{d}[gcd(x, y) = k]$

$ = sum_{x=a}^{b}(sum_{y=1}^{d}[gcd(x, y) = k] - sum_{y=1}^{c-1}[gcd(x, y) = k])$

$ = sum_{x=1}^{b}(sum_{y=1}^{d}[gcd(x, y) = k] - sum_{y=1}^{c-1}[gcd(x, y) = k]) - sum_{x=1}^{a - 1}(sum_{y=1}^{d}[gcd(x, y) = k] - sum_{y=1}^{c-1}[gcd(x, y) = k])$

$ = sum_{x=1}^{b}sum_{y=1}^{d}[gcd(x, y) = k] - sum_{x=1}^{b}sum_{y=1}^{c-1}[gcd(x, y) = k] - sum_{x=1}^{a-1}sum_{y=1}^{d}[gcd(x, y) = k] + sum_{x=1}^{a-1}sum_{y=1}^{c-1}[gcd(x, y) = k]$

令$sum(n, m) = sum_{x=1}^{n}sum_{y=1}^{m}[gcd(x, y) = k]$, 对$sum(n, m)$进行计算:

  $ sum_{x=1}^{n}sum_{y=1}^{m}[gcd(x, y) = k]$

$= sum_{x=1}^{lfloorfrac{n}{k} floor}sum_{y=1}^{lfloorfrac{m}{k} floor}[gcd(x, y) = 1]$

$ = sum_{x=1}^{lfloorfrac{n}{k} floor}sum_{y=1}^{lfloorfrac{m}{k} floor}varepsilon(gcd(i, j))$

$ = sum_{x=1}^{lfloorfrac{n}{k} floor}sum_{y=1}^{lfloorfrac{m}{k} floor}sum_{i|gcd(x, y)} mu(i)$

$ = sum_{i=1} mu(i) sum_{x=1}^{lfloorfrac{n}{k} floor}[i | x]sum_{y=1}^{lfloorfrac{m}{k} floor}[i | y]$

$ = sum_{i=1} lfloorfrac{n}{ki} floor lfloorfrac{m}{ki} floor$

即：$sum(n, m) = sum_{i=1} lfloorfrac{n}{ki} floor lfloorfrac{m}{ki} floor$

代入原式：$ans = sum(b, d) - sum(b, c - 1) - sum(a - 1, d) + sum(a - 1, c - 1)$即为答案

Code:

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
// #include <array>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
// #include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 1e7 + 20;
const int M = 1e5 + 20;
const int mod = 1e9 + 7;
const double eps = 1e-6;

ll cnt = 0, primes[N], phi[N], mu[N], sum[N] = {0};
map<ll, ll> su;
bool st[N];

int n, m, p, k;

void get(ll n){
    // phi[1] = 1;
    mu[1] = 1;
    for(ll i = 2; i <= n; i ++){
        if(!st[i]){
            primes[cnt ++] = i;
            // phi[i] = i - 1;
            mu[i] = -1;
        }
        for(int j = 0; primes[j] <= n / i; j ++){
            st[i * primes[j]] = true;
            if(i % primes[j] == 0){
                // phi[i * primes[j]] = primes[j] * phi[i];
                mu[i * primes[j]] = 0;
                break;
            }
            // phi[i * primes[j]] = (primes[j] - 1) * phi[i];
            mu[i * primes[j]] = - mu[i];
        }
    }

    for(int i = 1; i <= n; i ++){
        sum[i] = sum[i - 1] + mu[i];
    }
}

int get_sum(int n){
    if(n <= k) return sum[n];
    if(su[n]) return su[n];

    ll res = 1;
    for(int i = 2, r; i <= n; i = r + 1){
        r = n / (n / i);
        res -= (r - i + 1) * get_sum(n / i);
    }
    return su[n] = res;
}

int cal(int a, int b){
    int res = 0, mid1, mid2 = 0;
    if(a > b) swap(a, b);
    for(int i = 1, r; i <= a; i = r + 1){
        r = min(a / (a / i), b / (b / i));
        mid1 = get_sum(r);
        res = res + (mid1 - mid2) * (a / i) * (b / i);
        mid2 = mid1;
    }
    return res;
}

void solve()
{
    int a, b, c, d, kk;
    sdd(a, b);
    sdd(c, d);
    sd(kk);

    a --, c --;

    a /= kk, b /= kk, c /= kk, d /= kk;
    int ans = cal(b, d) - cal(b, c) - cal(a, d) + cal(a, c);
    cout << ans << endl;
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out.txt", "w", stdout);
    // freopen("/home/jungu/code/practice/out.txt","w",stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    sd(T);
    k = (int)pow(50000, 0.666667);
    get(k);
    // init(10000000);
    // int cas = 1;
    while (T--)
    {
        // printf("Case #%d:", cas++);
        solve();
    }

    return 0;
}