3518. 进化序列(evolve)

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题目：

传送门

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分析：

把所有数按位去做，通过位运算得出每位上 $1$的个数
然后再疯狂暴力判断是否符合条件，是就累加答案

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代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring> 
#include<cstdlib>
#include<algorithm>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<list>
#include<ctime>
#include<iomanip>
#include<string>
#include<bitset>
#include<deque>
#include<set>
#define LL long long
#define ch cheap
using namespace std;
inline LL read() {
    LL d=0,f=1;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
    return d*f;
}
LL a[100005],ans,v[35];
int main()
{
	freopen("evolve.in","r",stdin);
	freopen("evolve.out","w",stdout);
	LL n=read(),m=read();
	for(LL i=1;i<=n;i++) a[i]=read();
	LL last=1;
	for(LL i=1;i<=n;i++)
	{
		LL z=0;
		for(LL j=0;j<=30;j++)
		  v[j]+=(a[i]>>j)&1,z+=(v[j]!=0)<<j;
		while(z>=m&&last<i)
		{
			z=0;
			for(LL j=0;j<=30;j++)
			  v[j]-=(a[last]>>j)&1,z+=(v[j]!=0)<<j;
			last++;
		}
		ans+=i-last;
	}
	cout<<ans;
	return 0;
}