牛客小白月赛15 H 数据结构题___分块

题目大意：

一个长度为$n$的序列$a$，有$m$个询问，
每个询问给出$(l,r,l1,r1,x)$，
回答$get(l,r,x)\ast get(l1,r1,x)$ $mod$ $20180623$

$get(a,b,c)$表示区间$[a,b]$中有多少个$a_i$满足$a_i=c$

$1<=n,m<=10^5;1<=l,r,l1,r1<=10^5,a[i]\le 10000,x\le 100000$
$20180623不是质数$

分析：

分块，每个块内都存下这个块内的数，并将其排序，
每次查询就是枚举对应块，首尾直接枚举，中间的各个块二分即可，
时间复杂度：$O(m\ast \sqrt{n}\ast log\sqrt{n})$

代码：

#include 

#define N 100005
#define M 1005

using namespace std;

typedef long long ll;

const int mo = 20180623;

struct Node { int l, r, tot; ll c[M]; }d[M];
int n, m, cdp, cnt;
ll a[N];

int Work(int l, int r, ll x)
{
    if (l > r) swap(l, r);
    int sum = 0;
    int posl = l / cdp + (l % cdp ? 1 : 0);
    int posr = r / cdp + (r % cdp ? 1 : 0);
	if (posl == posr)    
    {
    	for (int i = l; i <= r; i++) sum += (a[i] == x ? 1 : 0);
    	return sum;
	}
	for (int i = l; i <= d[posl].r; i++) sum += (a[i] == x ? 1 : 0);
	for (int i = d[posr].l; i <= r; i++) sum += (a[i] == x ? 1 : 0);
	for (int i = posl + 1; i <= posr - 1; i++)
	{
		int ll = lower_bound(d[i].c + 1, d[i].c + d[i].tot + 1, x) - d[i].c;
		int rr = upper_bound(d[i].c + 1, d[i].c + d[i].tot + 1, x) - d[i].c - 1;
		if (d[i].c[ll] == x  && d[i].c[rr] == x) sum += (rr - ll + 1);  
	}
    return sum;
}

int main()
{
	scanf("%d %d", &n, &m);
	cdp = sqrt(n), cnt = 1, d[1].l = 1;
	for (int i = 1; i <= n; i++) 
	{
	    scanf("%lld", &a[i]);
		d[cnt].c[++d[cnt].tot] = a[i];
	    if (i == cdp * cnt) d[cnt++].r = i, d[cnt].l = i + 1; 
    }
    d[cnt].r = n; 
    for (int i = 1; i <= cnt; i++) sort(d[i].c + 1, d[i].c + d[i].tot + 1);
	int l1, r1, l2, r2; ll num;
	for (int i = 1; i <= m; i++)
	{
		scanf("%d %d %d %d %lld", &l1, &r1, &l2, &r2, &num);
		int num1 = Work(l1, r1, num);
		int num2 = Work(l2, r2, num);
		int ans = (ll)num1 * num2 % mo;
		printf("%d\n%d\n%d\n", num1, num2, ans);
	} 
	return 0;
}