AcWing 第59场周赛

https://www.acwing.com/activity/content/competition/problem_list/2015/

AcWing 4491. 数组操作

按题意模拟即可
注意小坑：前缀和包括第0项，多一个答案0

#include <bits/stdc++.h>

using namespace std;

int main () {
    int minn = 0, ans = 0, sum = 0;
    int n;  cin >> n;
    while (n --) {
        int x;
        cin >> x;
        ans += x, sum += x;
        minn = min (minn, sum);
    }
    cout << ans - minn;
}

AcWing 4492. 减法操作

对于给定的数n，有三种可能：

1. n为质数，那么最小质因子就是本身，所以答案为1；
2. n为偶数，最小质因子为2，减到0需要n/2次，答案为n/2
3. n为奇数，暴力枚举找到最小质因子p（必为奇数），此时数字变为\(n-p\)（奇数相加减得偶数），转化为情况2，所以答案为\(1+(n-p)/2\)

#include <bits/stdc++.h>
#define int long long

using namespace std;

bool is_prime (int x) {
    if (x == 2) return true;
    for (int i = 2; i*i <= x; i ++)
        if (x % i == 0) return false;
    return true;
}

signed main () {
    int n;
    cin >> n;
    //pre();
    if (is_prime (n))   cout << 1;
    else if (n % 2 == 0)    cout << n/2;
    else {
        for (int i = 2; i <= n; i ++) {
            if (is_prime (i) && n % i == 0) {
                cout << (n-i) / 2 + 1;
                break;
            }
        }

    }
}

4493. 环形连通分量

找连通块，判环
由无向图 + 环 得性质：所有点的度数都是2
所以只需枚举每个集合，判断该集合中是否所有点的度数都是2

#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 5;
int fa[N], d[N];
int n, m;
bool st[N]; //是否为环

int find (int x) {
    if (x != fa[x])
        fa[x] = find (fa[x]);
    return fa[x];
}

int main () {
    memset (st, true, sizeof st);
    cin >> n >> m;
    for (int i = 1; i <= n; i ++)   fa[i] = i;
    while (m --) {
        int a, b;
        cin >> a >> b;
        fa[find (a)] = find (b);
        d[a] ++, d[b] ++;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i ++)
        if (d[i] != 2)  st[find(i)] = false;

    for (int i = 1; i <= n; i ++)
        if (fa[i] == i && st[i])    cnt ++;

    cout << cnt << endl;
}

标签：周赛,59,int,fa,using,include,find,AcWing
来源： https://www.cnblogs.com/CTing/p/16462296.html