[SNOI2017]一个简单的询问

题目

写水题快乐一下

显然(operatorname{get(l_1,r_1,x)} imes operatorname{get(l_2,r_2,x)})可以拆成
(operatorname{pre(r_1,x)} imes operatorname{pre(r_2,x)}+operatorname{pre(l_1-1,x)} imes operatorname{pre(l_2-1,x)})减去(operatorname{pre(r_1,x)} imes operatorname{pre(l_2-1,x)}+operatorname{pre(r_2,x)} imes operatorname{pre(l_1-1,x)})

(operatorname{pre(i,x)})表示([1,i])中(x)出现次数

显然我们把一个询问拆成四个，(operatorname{pre(l,x)} imes operatorname{pre(r,x)})变成一组询问(l,r)即可，直接大力莫队莽

代码

#include <bits/stdc++.h>
#define re register
#define LL long long
inline int read() {
    char c = getchar();
    int x = 0;
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - 48, c = getchar();
    return x;
}
int a[50005], tax[50005], tmp[50005];
struct Ask {
    int l, r, o, rk;
} q[200005];
int n, m, Q, sz;
LL ans, Ans[50005];
inline int cmp(const Ask &A, const Ask &B) {
    if (A.l / sz == B.l / sz)
        return A.r < B.r;
    return A.l < B.l;
}
inline void del(int x, int o) {
    ans -= 1ll * tax[a[x]] * tmp[a[x]];
    tmp[a[x]] += o;
    ans += 1ll * tax[a[x]] * tmp[a[x]];
}
inline void add(int x, int o) {
    ans -= 1ll * tax[a[x]] * tmp[a[x]];
    tax[a[x]] += o;
    ans += 1ll * tax[a[x]] * tmp[a[x]];
}
int main() {
    n = read();
    for (re int i = 1; i <= n; i++) a[i] = read();
    Q = read();
    for (re int l1, l2, r1, r2, i = 1; i <= Q; i++) {
        l1 = read(), r1 = read(), l2 = read(), r2 = read();
        q[++m] = (Ask){ r1, r2, 1, i };
        if (l1 > 1 && l2 > 1)
            q[++m] = (Ask){ l1 - 1, l2 - 1, 1, i };
        if (l2 > 1)
            q[++m] = (Ask){ r1, l2 - 1, -1, i };
        if (l1 > 1)
            q[++m] = (Ask){ r2, l1 - 1, -1, i };
    }
    for (re int i = 1; i <= m; i++)
        if (q[i].l > q[i].r)
            std::swap(q[i].l, q[i].r);
    sz = n / (std::sqrt(m)) + 1;
    std::sort(q + 1, q + m + 1, cmp);
    for (re int l = 0, r = 0, i = 1; i <= m; i++) {
        while (r < q[i].r) add(++r, 1);
        while (l > q[i].l) del(l--, -1);
        while (l < q[i].l) del(++l, 1);
        while (r > q[i].r) add(r--, -1);
        Ans[q[i].rk] += 1ll * q[i].o * ans;
    }
    for (re int i = 1; i <= Q; i++) printf("%lld
", Ans[i]);
    return 0;
}