Codeforces Round #604 (Div. 2)

Posted 2021-03-20 swiftac

A.Beautiful String

分析：这个题扫一遍就好了，看看能不能放，放就完事了，最后检验一下是不是合法的。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+7;
char s[maxn];
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%s", s + 1);
        int n = strlen(s + 1);
        s[0] = ‘z‘;
        for (int i = 1; i <= n; i++) {
            if (s[i] == ‘?‘) {
                for (int c = ‘a‘; c <= ‘c‘; c++) {
                    if (c != s[i - 1] && c != s[i + 1]) {
                        s[i] = c;
                        break;
                    }
                }
            }
        }
        bool f = true;
        for (int i = 1; i <= n; i++) {
            if (s[i] == s[i + 1]) {
                f = false;
                break;
            }
        }
        if (f)
            printf("%s
", s + 1);
        else
            printf("-1
");
    }
    return 0;
}

B.Beautiful Numbers

分析：这个题找到1的位置，然后往两边延申一下弄一下就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+7;
int a[maxn];
int res[maxn];
priority_queue<int,vector<int>,less<int> > pq;
int main(){
    int t,n;
    cin>>t;
    while(t--) {
        while (!pq.empty()) pq.pop();
        cin >> n;
        for (int i = 1; i <= n; i++)
            cin >> a[i];
        int l = 0, r = 0;
        for (int i = 1; i <= n; i++) {
            if (a[i] == 1) {
                l = i, r = i;
                break;
            }
        }
        res[1] = 1;
        int cnt = 1;
        for (int i = 2; i <= n; i++) {
            if (l > 1 && r < n) {
                if (a[l - 1] < a[r + 1]) {
                    l--;
                    if (a[l] <= i) cnt++;
                    else pq.push(a[l]);
                } else {
                    r++;
                    if (a[r] <= i) cnt++;
                    else pq.push(a[r]);
                }
            } else {
                if (l <= 1) {
                    r++;
                    if (a[r] <= i) cnt++;
                    else pq.push(a[r]);
                } else {
                    l--;
                    if (a[l] <= i) cnt++;
                    else pq.push(a[l]);
                }
            }
            while (!pq.empty() && pq.top() <= i) {
                cnt++;
                pq.pop();
            }
            if (cnt == i) res[i] = 1;
            else res[i] = 0;
        }
        for (int i = 1; i <= n; i++)
            printf("%d", res[i]);
        printf("
");
    }
    return 0;
}

C.Beautiful Regional Contest

分析：先使得金牌尽可能地少，就是题量最多的那一堆人，让他们得金牌。然后往后再扫一扫让银牌尽可能地少，只要比金牌多然后合法就截止了，最后剩下的能让他得铜牌就得铜牌就好了。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=4e5+7;
int a[maxn],n;
int main() {
    int t;
    cin >> t;
    while (t--) {
        cin >> n;
        for (int i = 0; i < n; i++)
            cin >> a[i];
        int s = 1, g = 1, b = 0, i;
        for (i = 1; i < n / 2; i++) {
            if (a[i] != a[i - 1])
                break;
            g++;
        }
        for (i++; i < n / 2; i++) {
            if (s > g && a[i] != a[i - 1])
                break;
            s++;
        }
        int j = i - 1;
        while (i < n / 2) {
            while (i < n - 1 && a[i + 1] == a[i])
                i++;
            if (i < n / 2) b = i - j;
            i++;
        }
        if (s <= g || b <= g || !s || !g || !b) {
            printf("0 0 0
");
        } else {
            printf("%d %d %d
", g, s, b);
        }
    }
    return 0;
}