大意: 給定序列, 給定常數a,b, 兩種操作, (1)任選一個長為$t$的子區間刪除(不能全部刪除), 花費t*a. (2)任選$t$個元素+1/-1, 花費t*b. 求使整個序列gcd>1的最少花費.

題目有個限制是不能全部刪除, 所以最後一定剩余a[1]或a[n], 暴力枚舉a[1]與a[n]的所有素因子即可.

這場div. 2題目感覺都挺簡單的, 但實現起來各種出錯...........各種細節還是沒考慮好......

#include <iostream>
#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



const int N = 1e6+10;
int n, x, y;
int a[N], b[N];
ll ans;

ll solve(int g, int *a, int n) {
	int L=n+1,R=0,toty=0,totx=0;
	REP(i,0,n) b[i]=0;
	REP(i,1,n) if (a[i]%g) {
		if ((a[i]-1)%g&&(a[i]+1)%g) {
			L = min(L,i);
			R = i;
			++totx;
		}
		b[i]=1;
	}
	REP(i,L,R) b[i]=0;
	REP(i,1,n) toty+=b[i];
	totx = R-L+1;
	ll ans = (ll)totx*x+(ll)toty*y;
	if (L==n+1) {
		ans = (ll)toty*y;
		b[0] = 0;
		REP(i,1,n) b[i]+=b[i-1];
		ll mi = 0;
		REP(i,1,n) {
			ans = min(ans, (ll)toty*y+(ll)i*x-(ll)b[i]*y+mi);
			mi = min(mi, -(ll)i*x+(ll)b[i]*y);
		}
		return ans;
	}
	int dx = 0, dy = 0;
	ll d = 0;
	PER(i,1,L-1) {
		++dx;
		if (b[i]) --dy;
		d = min(d, (ll)dx*x+(ll)dy*y);
	}
	ans += d, d = 0, dx = dy = 0;
	REP(i,R+1,n) {
		++dx;
		if (b[i]) --dy;
		d = min(d, (ll)dx*x+(ll)dy*y);
	}
	ans += d;
	return ans;
}

vector<int> fac(int num) {
	int mx = sqrt(num+0.5);
	vector<int> v;
	REP(i,2,mx) if (num%i==0) {
		v.pb(i);
		while (num%i==0) num/=i;
	}
	if (num>1) v.pb(num);
	return v;
}

void solve(int num, int *a, int n, int tp) {
	vector<int> g = fac(num);
	for (auto &&t:g) ans = min(ans, solve(t,a,n)+tp*y);
}

int main() {
	scanf("%d%d%d", &n, &x, &y);
	REP(i,1,n) scanf("%d", a+i);
	ans = 1e15;
	solve(a[1],a+1,n-1,0);
	solve(a[1]-1,a+1,n-1,1);
	solve(a[1]+1,a+1,n-1,1);
	solve(a[n],a,n-1,0);
	solve(a[n]-1,a,n-1,1);
	solve(a[n]+1,a,n-1,1);
	printf("%lld\n", ans);
}

Array GCD CodeForces - 624D (數學,gcd)