Codeforces-446C. Pride

传送门

N（1~2000)个数，每次操作可以将相邻两数的其中一个变为它们的最大公约数，求将所有数变为1所需的最少操作次数

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;

const int maxn = 2e3 + 10;
int A[maxn];
int N;

int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}

int main() {
    int cnt = 0;
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) {
        scanf("%d", &A[i]);
        if (A[i] == 1) cnt++;
    }
    int ans = N + 1;
    for (int i = 1; i <= N; i++) {
        int a = 0;
        for (int j = i; j <= N; j++) {
            a = gcd(A[j], a);
            if (a == 1) {
                ans = min(ans, j - i + 1);
                break;
            }
        }
    }
    if (ans == N + 1) {
        puts("-1");
    } else {
        if (cnt) ans = N - cnt;
        else ans = N + ans - 2;
        printf("%d
", ans);
    }
    return 0;
}