different integers(j题)

事了拂衣去,深藏身与名

题目链接

  • 题意:区间内l,r的不同个数,每次查询只求新数

    莫队分块模板

Empirefree专属代码
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. **/


#include <algorithm>
#include <vector>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <queue>
#include <stdio.h>
using namespace std;

#define FFI(a, b) for(int i = a; i < b; i++)
#define FFJ(a, b) for(int j = a; j < b; j++)
#define FFK(a, b) for(int k = a; k < b; k++)
#define RR(a, b) for(int i = a; i > b; i++)
#define ME(a, b) memset(a, b, sizeof(a))
#define SC(x) scanf("%d", &x)
#define SCC(a, b) scanf("%d%d", &a, &b)
#define SCCC(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define PR(x) printf("%dn", x)
#define PRR(a, b) printf("%d%dn", a, b);
#define INF 0x3f3f3f3f
#define MAX 200010
#define MOD 1000000007
#define E 2.71828182845
#define M 8
typedef long long LL;
const double PI = acos(-1.0);
typedef pair<int, int> Author;
inline int () {
char c = getchar();int x = 0, f = 1;
while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
while(c >= '0' && c <= '9'){x = x * 10 + c - '0';c = getchar();}
return x * f;
}
/*******************************************************************************************************************************************************/
int pos[MAX], num[MAX], n, m, k, limit, l, r, s[MAX], ans;
struct Node{
int l;
int r;
int id;
int ans;
friend bool operator < (const Node& a, const Node& b){return pos[a.l] < pos[b.l] || (pos[a.l] == pos[b.l] && a.r < b.r);}
}ask[MAX];
inline bool cmp(const Node& a, const Node& b){return a.id < b.id;}
inline void modify(int p, int &ans, int add){
if(s[num[p]] == 1 && add == -1) ans--;
if(s[num[p]] == 0 && add == 1) ans++;
s[num[p]] += add;
}
int main(void){
while(~SCC(n, m)){
ME(s, 0);ans = 0;
limit = (int)sqrt(n * 2 + 0.5);
FFI(1, n + 1){num[i] = read();pos[i] = (i - 1) / limit + 1;}
FFI(1, n + 1){num[n + i] = num[i];pos[n + i] = (n + i - 1) / limit + 1;}
FFI(1, m + 1){ask[i].r = read();ask[i].l = read();ask[i].r += n; ask[i].id = i;}
sort(ask + 1, ask + m + 1);

l = 1; r = 0;
FFI(1, m + 1){
while(r < ask[i].r) modify(++r, ans, 1);
while(r > ask[i].r) modify(r--, ans, -1);
while(l < ask[i].l) modify(l++, ans, -1);
while(l > ask[i].l) modify(--l, ans, 1);
ask[i].ans = ans;
}
sort(ask + 1, ask + m + 1, cmp);
FFI(1, m + 1)PR(ask[i].ans);
}
return EXIT_SUCCESS;
}