BZOJ

题目链接：小Z的袜子

题意：$n$只袜子，$m$个询问，每次回答有多大概率在$[L,R]$区间内抽到两只颜色相同的袜子

思路：普通莫队，如果两个询问左端点在一个块内，则按询问右端点排序，否则按照所在块的序号排序，维护一个数组$num$，表示在区间$[L,R]$中，颜色为$c$的袜子有$num[c]$只，令变量$res=sum_{c}num[c]*(num[c]-1)/2$，显然对于每个区间$[L,R]$抽到同色袜子的概率就是$frac{res}{C_{R-L+1}^{2}}$，每次移动区间时修改$num[c]$，同时更新$res$即可，时间复杂度$O(n^{frac{3}{2}})$

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>

using namespace std;

typedef long long ll;

const int N = 50010;

struct node {
    int l, r, id;
};

int n, m, a[N];
int block, belong[N], tot, l[N], r[N];
ll num[N], res, mol[N], den[N];
node q[N];

void build()
{
    block = sqrt(n);
    tot = n / block;
    if (n % block) tot++;
    for (int i = 1; i <= tot; i++) {
        l[i] = (i - 1) * block + 1;
        r[i] = i * block;
    }
    r[tot] = n;
    for (int i = 1; i <= n; i++)
        belong[i] = (i - 1) / block + 1;
}

bool cmp(node a, node b)
{
    if (belong[a.l] != belong[b.l])
        return belong[a.l] < belong[b.l];
    return a.r < b.r;
}

void add(int x)
{
    res = res - num[a[x]] * (num[a[x]] - 1) / 2;
    num[a[x]]++;
    res = res + num[a[x]] * (num[a[x]] - 1) / 2;
}

void sub(int x)
{
    res = res - num[a[x]] * (num[a[x]] - 1) / 2;
    num[a[x]]--;
    res = res + num[a[x]] * (num[a[x]] - 1) / 2;
}

ll c(int n)
{
    return (ll)n * (n - 1) / 2;
}

ll gcd(ll a, ll b)
{
    return 0 == b ? a : gcd(b, a % b);
}

int main()
{
    scanf("%d%d", &n, &m);
    build();
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &q[i].l, &q[i].r);
        den[i] = c(q[i].r - q[i].l + 1);
        q[i].id = i;
    }
    sort(q + 1, q + m + 1, cmp);
    int L = 1, R = 0;
    for (int i = 1; i <= m; i++) {
        while (q[i].l < L) add(--L);
        while (q[i].r > R) add(++R);
        while (q[i].l > L) sub(L++);
        while (q[i].r < R) sub(R--);
        mol[q[i].id] = res;
    }
    for (int i = 1; i <= m; i++) {
        ll d = gcd(mol[i], den[i]);
        if (0 == mol[i]) printf("0/1
");
        else printf("%lld/%lld
", mol[i] / d, den[i] / d);
    }
    return 0;
}