题目链接

Solution CF220B Little Elephant and Array

题目大意：给定一个序列，多次询问在 \([l,r]\) 内，有多少个数 \(x\)，其出现次数也为 \(x\)。

暴力，奇妙的复杂度

分析：

首先可能成为答案的数字个数是 \(O(\sqrt{n})\) 的，然后大于 \(n\) 的数也是不可能成为答案的。因此我们直接暴力统计。

开个数组统计 \(x \leq n\) 的 \(x\) 的出现次数，记为 \(cnt[x]\)，只有 \(x \leq cnt[x]\) 的 \(x\) 才可能为答案，离线一下就可以了。

#include <cstdio>
#include <cctype>
#pragma GCC optmize(2)
using namespace std;
typedef long long ll;
constexpr int maxn = 1e5 + 100,inf = 0x7fffffff;
struct IO{//-std=c++11,with cstdio and cctype
	private:
		static constexpr int ibufsiz = 1 << 20;
		char ibuf[ibufsiz + 1],*inow = ibuf,*ied = ibuf;
		static constexpr int obufsiz = 1 << 20;
		char obuf[obufsiz + 1],*onow = obuf;
		const char *oed = obuf + obufsiz;
	public:
		inline char getchar(){
			#ifndef ONLINE_JUDGE
				return ::getchar();
			#else
				if(inow == ied){
					ied = ibuf + sizeof(char) * fread(ibuf,sizeof(char),ibufsiz,stdin);
					*ied = '\0';
					inow = ibuf;
				}
				return *inow++;
			#endif
		}
		template<typename T>
		inline void read(T &x){
			static bool flg;flg = 0;
			x = 0;char c = getchar();
			while(!isdigit(c))flg = c == '-' ? 1 : flg,c = getchar();
			while(isdigit(c))x = x * 10 + c - '0',c = getchar();
			if(flg)x = -x;
		}
		template <typename T,typename ...Y>
		inline void read(T &x,Y&... X){read(x);read(X...);}
		inline int readi(){static int res;read(res);return res;}
		inline long long readll(){static long long res;read(res);return res;}
		
		inline void flush(){
			fwrite(obuf,sizeof(char),onow - obuf,stdout);
			fflush(stdout);
			onow = obuf;
		}
		inline void putchar(char c){
			#ifndef ONLINE_JUDGE
				::putchar(c);
			#else
				*onow++ = c;
				if(onow == oed){
					fwrite(obuf,sizeof(char),obufsiz,stdout);
					onow = obuf;
				}
			#endif
		}
		template <typename T>
		inline void write(T x,char split = '\0'){
			static unsigned char buf[64];
			if(x < 0)putchar('-'),x = -x;
			int p = 0;
			do{
				buf[++p] = x % 10;
				x /= 10;
			}while(x);
			for(int i = p;i >= 1;i--)putchar(buf[i] + '0');
			if(split != '\0')putchar(split);
		}
		inline void lf(){putchar('\n');}
		~IO(){
			fwrite(obuf,sizeof(char),onow - obuf,stdout);
		}
}io;
template <typename A,typename B>
inline void chkmin(A &x,const B &y){if(y < x)x = y;}
template <typename A,typename B>
inline void chkmax(A &x,const B &y){if(y > x)x = y;}

int n,m,val[maxn],cnt[maxn],sum[maxn],ans[maxn],l[maxn],r[maxn];
int main(){
#ifndef ONLINE_JUDGE
	freopen("fafa.in","r",stdin);
#endif
	io.read(n,m);
	for(int i = 1;i <= n;i++){
		io.read(val[i]);
		if(val[i] <= n)cnt[val[i]]++;
	}
	for(int i = 1;i <= m;i++)io.read(l[i],r[i]);	
	for(int i = 1;i <= n;i++)
		if(i <= cnt[i]){
			for(int k = 1;k <= n;k++)sum[k] = sum[k - 1] + (i == val[k]);
			for(int k = 1;k <= m;k++)
				if(sum[r[k]] - sum[l[k] - 1] == i)ans[k]++;
		}
	for(int i = 1;i <= m;i++)io.write(ans[i],'\n');
	return 0;
}

标签：Little,int,题解,void,read,CF220B,maxn,inline,obuf
来源： https://www.cnblogs.com/colazcy/p/14084375.html