Codeforces Educational Codeforces Round 67

Contest Info

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Data：2019.6.30
Solved：4/7

Solutions

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A. Stickers and Toys

Meaning of the questions:
There \ (A \) Items \ (s \) a, \ (B \) Items \ (t \) months, these items will now be loaded into the \ (n \) boxes, each box is only about three situations:

• Only a \ (A \) items
• Only a \ (B \) items
• A \ (A \) articles and a \ (B \) Item

Now you ask, how many boxes you want to take at least can guarantee you at least have a \ (A \) items and a \ (B \) items.

Ideas:
The pigeonhole principle, be apparent to \ (A \) article, taking at least \ (n - s + 1 \ ) a box can have a \ (A \) items.
Similarly, for the \ (B \) article to take at least \ (n - t + 1 \ ) boxes.
The answer is \ (Min (n - s +1 , n - t + 1) \)

Code:

#include <bits/stdc++.h>
using namespace std;
 
int main() {
    int n, s, t;
    int T; scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d", &n, &s, &t);
        int res = max(n - s + 1, n - t + 1);
        printf("%d\n", res);
    }
    return 0;
}

B. Letters Shop

The meaning of problems:
there is a string \ (s \) , Always ask a string \ (t \) , ask a minimum of \ (s \) prefix character makes this prefix may have composed a string \ (t \) .

A thought:
can maintain a prefix and the number of characters, then two.

Code One:

#include <bits/stdc++.h>
using namespace std;
 
#define N 200010
int n, m, lens, lent;
char s[N], t[N];
int sum[N][27];
int cnt[27];
 
bool ok(int x) {
    for (int i = 0; i < 26; ++i) {
        if (sum[x][i] < cnt[i]) {
            return 0;
        }
    }
    return 1;
}
 
int main() {
    while (scanf("%d", &n) != EOF) {
        memset(sum, 0, sizeof sum);
        scanf("%s", s + 1); lens = strlen(s + 1);
        for (int i = 1; i <= lens; ++i) {
            ++sum[i][s[i] - 'a'];
            for (int j = 0; j < 26; ++j) {
                sum[i][j] += sum[i - 1][j];
            }
        }
        scanf("%d", &m);
        while (m--) {
            scanf("%s", t + 1);  lent = strlen(t + 1);
            memset(cnt, 0, sizeof cnt);
            for (int i = 1; i <= lent; ++i) {
                ++cnt[t[i] - 'a'];
            }
            int l = 1, r = n, res = -1;
            while (r - l >= 0) {
                int mid = (l + r) >> 1;
                if (ok(mid)) {
                    r = mid - 1;
                    res = mid;
                } else {
                    l = mid + 1;
                }
            }
            printf("%d\n", res);
        }
    }
    return 0;
}

Thinking II:
Maintenance \ (S \) string class of the characters of \ (I \) th location, will be apparent to \ (T \) string for each type of character has \ (X \) th words, \ (S \) length greater than or equal to the prefix string of such character \ () X \ a position where a.

Code II:

#include <bits/stdc++.h>
using namespace std;
 
#define N 200010
int n, m, lens, lent;
char s[N], t[N];
int sum[N][27];
int cnt[27];
 
bool ok(int x) {
    for (int i = 0; i < 26; ++i) {
        if (sum[x][i] < cnt[i]) {
            return 0;
        }
    }
    return 1;
}
 
int main() {
    while (scanf("%d", &n) != EOF) {
        memset(sum, 0, sizeof sum);
        scanf("%s", s + 1); lens = strlen(s + 1);
        for (int i = 1; i <= lens; ++i) {
            ++sum[i][s[i] - 'a'];
            for (int j = 0; j < 26; ++j) {
                sum[i][j] += sum[i - 1][j];
            }
        }
        scanf("%d", &m);
        while (m--) {
            scanf("%s", t + 1);  lent = strlen(t + 1);
            memset(cnt, 0, sizeof cnt);
            for (int i = 1; i <= lent; ++i) {
                ++cnt[t[i] - 'a'];
            }
            int l = 1, r = n, res = -1;
            while (r - l >= 0) {
                int mid = (l + r) >> 1;
                if (ok(mid)) {
                    r = mid - 1;
                    res = mid;
                } else {
                    l = mid + 1;
                }
            }
            printf("%d\n", res);
        }
    }
    return 0;
}

C. Vasya And Array

Meaning of the questions: