BZOJ 2038 小Z的袜子(hose) (莫队算法)

题意：中文题。

析：很著名的莫队算法，先把这个求概率的式子表达出来，应该是分子：C(x1, 2) + C(x2, 2) + C(x3, 2) + ... + C(xn, 2)  分母：C(n, 2)，然后化成分数的表达形式，[x1(x1-1)+x2(x2-1)+...+xn(xn-1)] / (n*(n-1))  然后再化简得到 (sigma(xi*xi)  - n) / (n*(n-1)) ，然后就是对每个区间进行运算，离线，把所以的序列分成sqrt(n)块，然后用两个指针，进行对数据的计算。

注意不要用I64d，用的话WA到死。。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <assert.h>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 0xffffffffffLL;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int maxn = 50000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

LL ansx[maxn], ansy[maxn];
int pos[maxn], val[maxn];
int cnt[maxn];

struct Node{
  int l, r, id;
  bool operator < (const Node &p) const{
    return pos[l] < pos[p.l] || pos[l] == pos[p.l] && r < p.r;
  }
};

Node a[maxn];

LL x, y;

void update(int l, int ok){
  x -= cnt[val[l]] * (LL)cnt[val[l]];
  cnt[val[l]] += ok;
  x += cnt[val[l]] * (LL)cnt[val[l]] - ok;
  y += ok;
}

void update(int i){
  ansx[i] = x;
  ansy[i] = y * (y - 1);
}

int main(){
  scanf("%d %d", &n, &m);
  for(int i = 1; i <= n; ++i) scanf("%d", val+i);
  for(int i = 0; i < m; ++i){
    scanf("%d %d", &a[i].l, &a[i].r);
    a[i].id = i;
  }
  int t = sqrt(n + 0.5);
  for(int i = 1; i <= n; ++i)
    pos[i] = i / t;
  sort(a, a + m);
  for(int l = 1, i = 0 ,r = 0; i < m; ++i){
    int L = a[i].l, R = a[i].r;
    while(l < L)  update(l++, -1);
    while(l > L)  update(--l, 1);
    while(r < R)  update(++r, 1);
    while(r > R)  update(r--, -1);
    update(a[i].id);
  }
  for(int i = 0; i < m; ++i){
    LL g = gcd(ansx[i], ansy[i]);
    printf("%lld/%lld
", ansx[i]/g, ansy[i]/g);
  }
  return 0;
}