Educational Codeforces Round 53 (Rated for Div. 2) A，B，C，D，E题解

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A. Diverse Substring

题意：找一个子串满足任何一个字符的出现次数小于 n/2 n是子串的长度，

显然两个不相同的字符组成的子串就是满足情况的；

#include<bits/stdc++.h>

using namespace std;

#define out fflush(stdout)
#define fast ios::sync_with_stdio(0),cin.tie(0);

#define FI first
#define SE second

typedef long long ll;
typedef pair<int,int> P;

const int maxn = 1000 + 7;
const int INF = 0x3f3f3f3f;


int n;
char s[maxn];
map<char, int> mp;

int main() {
    scanf("%d", &n);
    scanf("%s", s+1);

    for(int i = 2; i <= n; ++i) {
        if(s[i] != s[i-1]) {
            puts("YES");
            printf("%c%c", s[i-1], s[i]);
            return 0;
        }
    }
    puts("NO");
    return 0;
}

B. Vasya and Books

题意：给定n本书的位置，从上往下，编号1~n不重复，没找一本书需要把他上面的书全搬走，问每找一本书需要搬几本书，如果之前已经搬出来的书不用搬动，输出0；

思路：直接模拟

#include<bits/stdc++.h>

using namespace std;

#define out fflush(stdout)
#define fast ios::sync_with_stdio(0),cin.tie(0);

#define FI first
#define SE second

typedef long long ll;
typedef pair<int,int> P;

const int maxn = 2e5 + 7;
const int INF = 0x3f3f3f3f;


int n;
int a[maxn], b[maxn];

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &a[i]);
    }
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &b[i]);
    }
    int id = 1;
    set<int> st;
    for(int i = 1; i <= n; ++i) {
        if(st.count(b[i])) {
            printf("%d%c", 0, (i == n ? '\n' : ' '));
        }
        else {
            int ans = 1;
            while(a[id] != b[i]) {
                st.insert(a[id]);
                id++, ans++;
            }
            id++;
            printf("%d%c", ans, (i == n ? '\n' : ' '));
        }
    }


    return 0;
}

C. Vasya and Robot

题解：https://blog.csdn.net/xiang_6/article/details/83686299

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D. Berland Fair

题解：https://blog.csdn.net/xiang_6/article/details/83686379

E. Segment Sum

题解：