sjtu1364 countcountcount
Description
我有一个元素个数为\(n\)的整数数组\(a\)和\(Q\)个问题,每个问题有\(x,y\)两个参数,我要数有多少个整数\(K\)满足\(K\)在\(a[x]…a[y]\)中出现了恰好\(K\)次。
题面就这么简单了。
Input
第一行两个整数\(n,Q\),表示数组\(a\)的元素个数和询问数;
接下来一行n给整数,描述数组a;
接下来Q行,每行两个数xi,yi(1<=xi<=yi<=n),表示询问的左右边界;
Output
输出Q行,每行一个整数表示满足询问的K的个数。
Sample Input
7 2
3 1 2 2 3 3 7
1 7
3 4
Sample Output
3
1
Hint
对于30%的数据,\(1 \le n,Q \le 1000\);
对于100%的数据,\(1 \le n,Q \le 100000,1 \le a[i] \le 10^9\);
sb莫队题目。。。1个元素支持\(O(1)\)加入。
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
using namespace std;
#define maxn (100010)
int N,tot,Q,A[maxn],num[maxn],ans[maxn],lim,size;
inline int gi()
{
char ch; int f = 1,ret = 0;
do ch = getchar(); while (!(ch >= '0'&&ch <= '9')&&ch != '-');
if (ch == '-') f = -1,ch = getchar();
do ret = ret*10+ch-'0',ch = getchar(); while (ch >= '0'&&ch <= '9');
return ret*f;
}
struct node
{
int l,r,id;
inline void read(int i) { l = gi(),r = gi(); id = i; lim = max(lim,r); }
}query[maxn];
inline bool cmp1(const node &a,const node &b) { return a.l < b.l; }
inline bool cmp2(const node &a,const node &b) { return a.r < b.r; }
inline void work()
{
int res = 0,L = query[1].l,R = query[1].r;
for (int i = L;i <= R;++i)
if (A[i] <= N)
{
if (num[A[i]] == A[i]) --res;
if (++num[A[i]] == A[i]) ++res;
}
for (int i = 1;i <= Q;++i)
{
while (R < query[i].r)
{
++R;
if (A[R] <= N)
{
if (num[A[R]] == A[R]) --res;
if (++num[A[R]] == A[R]) ++res;
}
}
while (R > query[i].r)
{
if (A[R] <= N)
{
if (num[A[R]] == A[R]) --res;
if (--num[A[R]] == A[R]) ++res;
}
--R;
}
while (L > query[i].l)
{
--L;
if (A[L] <= N)
{
if (num[A[L]] == A[L]) --res;
if (++num[A[L]] == A[L]) ++res;
}
}
while (L < query[i].l)
{
if (A[L] <= N)
{
if (num[A[L]] == A[L]) --res;
if (--num[A[L]] == A[L]) ++res;
}
++L;
}
ans[query[i].id] = res;
}
}
int main()
{
freopen("1364.in","r",stdin);
freopen("1364.out","w",stdout);
N = gi(); Q = gi();
for (int i = 1;i <= N;++i) A[i] = gi();
for (int i = 1;i <= Q;++i) query[i].read(i);
size = ceil(sqrt(lim)+0.5);
sort(query+1,query+Q+1,cmp1);
for (int i = 1,j;i <= Q;i = j+1)
{
for (j = i;j < Q&&query[j+1].l - query[i].l <= size;++j);
sort(query+i,query+j+1,cmp2);
}
work();
for (int i = 1;i <= Q;++i) printf("%d\n",ans[i]);
fclose(stdin); fclose(stdout);
return 0;
}
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