洛谷P1494 小Z的袜子

题意：在[l, r]之中任选两个数，求它们相同的概率。

解：

莫队入门。

概率这个很好搞，就是cnt * (cnt - 1) / 2。

然后发现每次挪指针的时候，某一个cnt会+1或-1。这时候差值就是2 * cnt_small。

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

typedef long long LL;
const int N = 50010;

int a[N], bin[N], fr[N];
LL ans;

struct ASK {
    int id, l, r;
    LL ans;
    inline bool operator <(const ASK &w) const {
        if(fr[l] != fr[w.l]) {
            return fr[l] < fr[w.l];
        }
        return r < w.r;
    }
}ask[N];

inline bool cmp(const ASK &a, const ASK &b) {
    return a.id < b.id;
}

inline LL gcd(LL a, LL b) {
    if(!b) {
        return a;
    }
    return gcd(b, a % b);
}

inline void add(int x) {
    ans += bin[a[x]] << 1;
    bin[a[x]]++;
    return;
}

inline void del(int x) {
    bin[a[x]]--;
    ans -= bin[a[x]] << 1;
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    int T = sqrt(n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        fr[i] = (i - 1) / T + 1;
    }
    for(int i = 1; i <= m; i++) {
        ask[i].id = i;
        scanf("%d%d", &ask[i].l, &ask[i].r);
    }
    std::sort(ask + 1, ask + m + 1);

    int L = 1, R = 1;
    bin[a[1]]++;
    for(int i = 1; i <= m; i++) {
        if(ask[i].l == ask[i].r) {
            continue;
        }
        while(ask[i].l < L) {
            add(--L);
        }
        while(R < ask[i].r) {
            add(++R);
        }
        while(L < ask[i].l) {
            del(L++);
        }
        while(ask[i].r < R) {
            del(R--);
        }
        ask[i].ans = ans;
    }

    std::sort(ask + 1, ask + m + 1, cmp);
    for(int i = 1; i <= m; i++) {
        if(!ask[i].ans || ask[i].l == ask[i].r) {
            puts("0/1");
            continue;
        }
        int g = gcd(ask[i].ans, 1ll * (ask[i].r - ask[i].l + 1) * (ask[i].r - ask[i].l));
        printf("%lld/%lld
", ask[i].ans / g, 1ll * (ask[i].r - ask[i].l + 1) * (ask[i].r - ask[i].l) / g);
    }
    return 0;
}