【BZOJ】3289: Mato的文件管理（莫队算法+树状数组）

#include <cstdio>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }

const int N=50005;
struct dat { int l, r, id; }q[N];
int a[N], c[N], m, n, t[N], ans[N], sum, tot, pos[N];

void add(int x, int s) { for(; x<=tot; x+=x&-x) c[x]+=s; }
int getsum(int x) { int r=0; for(; x; x-=x&-x) r+=c[x]; return r; }
bool cmp(const dat &a, const dat &b) { return pos[a.l]==pos[b.l]?a.r<b.r:a.l<b.l; }

void fix(int x, int f, int fx, int len) { //0:left 1:right
	int y=(fx?len-getsum(x):getsum(x-1));
	sum+=f*y;
	add(x, f);
}

void init() {
	int sz=sqrt(0.5+n);
	for1(i, 1, n) pos[i]=i/sz;
	sort(q+1, q+1+m, cmp);
	//for1(i, 1, m) dbg(q[i].id);
}

int main() {
	read(n);
	for1(i, 1, n) read(a[i]), t[i]=a[i];
	sort(t+1, t+1+n);
	tot=unique(t+1, t+1+n)-t-1;
	for1(i, 1, n) a[i]=lower_bound(t+1, t+1+tot, a[i])-t;
	read(m);
	for1(i, 1, m) read(q[i].l), read(q[i].r), q[i].id=i;
	init();
	int xl, xr, id, l=1, r=0;
	for1(i, 1, m) {
		xl=q[i].l;
		xr=q[i].r;
		id=q[i].id;
		while(l<xl) fix(a[l], -1, 0, r-l+1), ++l;
		while(l>xl) fix(a[l-1], 1, 0, r-l+1), --l;
		while(r<xr) fix(a[r+1], 1, 1, r-l+1), ++r;
		while(r>xr) fix(a[r], -1, 1, r-l+1), --r;
		ans[id]=sum;
	}
	for1(i, 1, m) printf("%d\n", ans[i]);

	return 0;
}