CodeForces Round #529 Div.3

http://codeforces.com/contest/1095

A. Repeating Cipher

#include <bits/stdc++.h>
using namespace std;

int N;
string s;
string ans = "";

int main() {
    scanf("%d", &N);
    cin >> s;
    int cnt = 0;
    for(int i = 0; i < N;) {
        ans += s[i];
        cnt ++;
        i += cnt;
    }
    cout << ans;
    return 0;
}

B. Array Stabilization

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N;
int a[maxn];

int main() {
    scanf("%d", &N);
    for(int i = 0; i < N; i ++)
        scanf("%d", &a[i]);

    sort(a, a + N);
    int ans1 = a[N - 2] - a[0];
    int ans2 = a[N - 1] - a[1];
    printf("%d
", min(ans1, ans2));
    return 0;
}

C. Powers Of Two

#include <bits/stdc++.h>
using namespace std;

int N, K;
priority_queue<int> q;
int cnt = 0;

int main() {
    scanf("%d%d", &N, &K);
    for(int i = 30; i >= 0; i --) {
        if(N >= (1 << i)) {
            q.push(i);
            N -= (1 << i);
            cnt ++;
        }
    }

    if(K < cnt) printf("NO
");
    else {
        while(cnt < K) {
            int rec = q.top();
            if(rec == 0) {
                printf("NO
");
                return 0;
            }

            q.pop();
            q.push(rec - 1);
            q.push(rec - 1);

            cnt ++;
        }

        printf("YES
");
        while(!q.empty()) {
            int t = q.top();
            printf("%d ", (1 << t));
            q.pop();
        }

    }

    return 0;
}

D. Circular Dance

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + 10;
vector<int> v[maxn];
int a[maxn][3], vis[maxn];
int N;
vector<int> ans;

void dfs(int step) {
    vis[step] = 1;
    ans.push_back(step);
    for(int i = 0; i < v[step].size(); i ++) {
        if(!vis[v[step][i]] && (v[step][i] == a[step][0] || v[step][i] == a[step][1])) {
            vis[v[step][i]] = 1;
            dfs(v[step][i]);
        }
    }
}

int main() {
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++) {
        int st, en;
        scanf("%d%d", &st, &en);
        a[i][0] = st, a[i][1] = en;
        v[st].push_back(en);
        v[en].push_back(st);
    }

    dfs(1);
    for(int i = 0; i < ans.size(); i ++)
        printf("%d%s", ans[i], i != ans.size() - 1 ? " " : "
");

    return 0;
}

E. Almost Regular Bracket Sequence

（合法的括号匹配串的充要条件是 ① 前 i 项 前缀和非负 ② sum[N] == 0）

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e6 + 10;
int N;
string s;
int a[maxn], sum[maxn], b[maxn], c[maxn];

int main() {
    cin >> N >> s;
    for(int i = 0; i < N; i ++) {
        if(s[i] == '(')
            a[i] = 1;
        else a[i] = -1;
    }

    sum[0] = a[0];
    for(int i = 1; i < N; i ++)
        sum[i] = a[i] + sum[i - 1];

    c[0] = sum[0];
    for(int i = 1; i < N; i ++)
        c[i] = min(c[i - 1], sum[i]);

    b[N - 1] = sum[N - 1];
    for(int i = N - 2; i >= 0; i --)
        b[i] = min(sum[i], b[i + 1]);

    int ans = 0;
    for(int i = 0; i < N; i ++) {
        if(s[i] == '(') {
            if(c[i - 1] >= 0 && b[i] - 2 >= 0 && sum[N - 1] - 2 == 0) ans ++;
        } else {
            if(c[i - 1] >= 0 && b[i] + 2 >= 0 && sum[N - 1] + 2 == 0) ans ++;
        }
    }

    printf("%d
", ans);

    return 0;
}

F. Make It Connected

（很久没写最小生成树 这个先留一哈）

写的好烦 脑子不好用 想不到 然后开始吃薯片 还是暴躁 越来越暴躁 在放弃的边缘大鹏展翅 所以完全不想看题目 想豁奶茶想吹风