bzoj 2038 小Z的袜子

好久没写题解了=_= ，整个暑假就没写过，还是决定写写吧，所以挑了这道大水题。

这是标准的莫队算法的问题，但由于可能数据水还是别的什么原因，不用曼哈顿最小生成树也可以过。具体就是按询问区间的左端点分块， 块内按右端点排序，然后暴力……

真的是暴力，太暴力了，直到AC以后我才相信这么暴力真的可以在O(N^1.5)的时间复杂度内过掉。

块内具体就是右端点递增，左端点由于在块内并不是有序的，所以左端点就会晃来晃去，真是太暴力了……

上代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#define N 50001
using namespace std;

int n, m, kuai;
struct sss
{
    int l, r;
    int num, ku;
}ask[N];
struct ss
{
    long long first, second;
}ans[N];
int colorcount[N], color[N];

bool cmp(sss x, sss y) { return x.ku == y.ku ? x.r < y.r : x.ku < y.ku; }

long long gcd(long long x, long long y)
{
    if (!y) return x;
    return gcd(y, x%y);
}

int main()
{
    scanf("%d%d", &n, &m); kuai = (int)sqrt(n+0.5);
    for (int i = 1; i <= n; ++i) scanf("%d", &color[i]);
    for (int i = 1; i <= m; ++i)
    {
        scanf("%d%d", &ask[i].l, &ask[i].r);
        ask[i].num = i; ask[i].ku = ask[i].l / kuai + 1;
    }
    sort(ask+1, ask+1+m, cmp);
    int nowkuai = 0, nowr, nowl; long long nowans;
    for (int i = 1; i <= m; ++i)
    {
        if (ask[i].ku != nowkuai)
        {
            nowkuai = ask[i].ku; nowans = 0;
            nowl = ask[i].l; nowr = nowl-1;
            for (int j = 1; j <= n; ++j) colorcount[j] = 0;
        }
        for (int j = nowl; j < ask[i].l; ++j)
        {
            colorcount[color[j]] --;
            nowans -= colorcount[color[j]];
        }
        for (int j = nowl-1; j >= ask[i].l; --j)
        {
            colorcount[color[j]] ++;
            nowans += colorcount[color[j]] - 1;
        }
        for (int j = nowr+1; j <= ask[i].r; ++j)
        {
            colorcount[color[j]] ++;
            nowans += colorcount[color[j]] - 1;
        }
        nowl = ask[i].l; nowr = ask[i].r;
        long long x = ask[i].r - ask[i].l + 1;
        x = (long long)x*(long long)(x-1)/2;
        long long d = gcd(x, nowans);
        if (nowans)
        {
        ans[ask[i].num].first = nowans/d;
        ans[ask[i].num].second = x/d;
        }
        else
        {
            ans[ask[i].num].first = 0;
            ans[ask[i].num].second = 1;
        }
    }
    for (int i = 1; i <= m; ++i)
        printf("%lld/%lld\n", ans[i].first, ans[i].second);
    return 0;
}