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;
#endifint 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);
}
