Card Game Again CodeForces - 818E (双指针)

时间：2022-04-30 07:09:02

大意: 给定序列, 求多少个区间积被k整除.

整除信息满足单调性, 显然双指针. 具体实现只需要考虑k的素数向量, 对每一维维护个指针即可.

这题看了下cf其他人的做法, 发现可以直接暴力, 若当前的前缀积模k为0, 暴力向前求出第一个后缀积为0的位置即可, 复杂度是$O(n)$的并且相当好写.

#include <iostream>

#include <algorithm>

#include <cstdio>

#include <math.h>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <string.h>

#include <bitset>

#define REP(i,a,n) for(int i=a;i<=n;++i)

#define PER(i,a,n) for(int i=n;i>=a;--i)

#define hr putchar(10)

#define pb push_back

#define lc (o<<1)

#define rc (lc|1)

#define mid ((l+r)>>1)

#define ls lc,l,mid

#define rs rc,mid+1,r

#define x first

#define y second

#define io std::ios::sync_with_stdio(false)

#define endl '\n'

#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})

using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

const int P = 1e9+7, INF = 0x3f3f3f3f;

ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}

ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}

inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}

//head

#ifdef ONLINE_JUDGE

const int N = 1e6+10;

#else

const int N = 111;

#endif

int n, k;

int p[11], f[11], cnt;

int g[N][11], cur[11], sum[11];

int main() {

	scanf("%d%d", &n, &k);

	int mx = sqrt(k+0.5);

	REP(i,2,mx) if (k%i==0) {

		p[++cnt] = i;

		while (k%i==0) k/=i,++f[cnt];

	}

	if (k>1) p[++cnt]=k,++f[cnt];

	REP(j,1,n) {

		scanf("%d", &k);

		REP(i,1,cnt) if (k%p[i]==0) {

			while (k%p[i]==0) ++g[j][i],k/=p[i];

		}

	}

	ll ans = 0;

	int now = 0;

	REP(i,1,n) {

		REP(j,1,cnt) {

			while (cur[j]<n&&sum[j]<f[j]) sum[j]+=g[++cur[j]][j];

			if (sum[j]<f[j]) {

				printf("%lld\n", ans);

				return 0;

			}

			now = max(now, cur[j]);

		}

		now = max(now, i);

		ans += n-now+1;

		REP(j,1,cnt) sum[j]-=g[i][j];

	}

	printf("%lld\n", ans);

}

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