2021牛客暑期多校训练营4 C、LCS

题目大意

你需要构造个长度为的字符串，并且保证。

Solution

考点：模拟

我们先不考虑输出顺序，只考虑是否能够构建，我们把变成的形式进行一次排序。

那么我们可以想到下图这样的构建方式是最合理的：

接下来就是找出最小的字母，然后确定这三个字符串的位置了。

int solve() {
    string s1, s2, s3;
    int a = read(), b = read(), c = read(), n = read();

    int maxi = max({ a,b,c });
    int mini = min({ a,b,c });
    int tmp = a ^ b ^ c ^ maxi ^ mini;

    if (maxi + tmp - mini > n) {
        return puts("NO"), 0;
    }

    if (mini == b) {
        for (int i = 1; i <= b; ++i) {
            s1 += 'a', s2 += 'a', s3 += 'a';
        }
        for (int i = 1; i <= a - b; ++i) {
            s1 += 'b', s2 += 'b';
        }
        for (int i = 1; i <= c - b; ++i) {
            s1 += 'c', s3 += 'c';
        }
        while (s1.size() < n)    s1 += 'd';
        while (s2.size() < n)    s2 += 'e';
        while (s3.size() < n)    s3 += 'f';
    }
    else if (mini == c) {
        for (int i = 1; i <= c; ++i) {
            s1 += 'a', s2 += 'a', s3 += 'a';
        }
        for (int i = 1; i <= a - c; ++i) {
            s1 += 'b', s2 += 'b';
        }
        for (int i = 1; i <= b - c; ++i) {
            s2 += 'c', s3 += 'c';
        }
        while (s1.size() < n)    s1 += 'd';
        while (s2.size() < n)    s2 += 'e';
        while (s3.size() < n)    s3 += 'f';
    }
    else if (mini == a) {
        for (int i = 1; i <= a; ++i) {
            s1 += 'a', s2 += 'a', s3 += 'a';
        }
        for (int i = 1; i <= b - a; ++i) {
            s2 += 'b', s3 += 'b';
        }
        for (int i = 1; i <= c - a; ++i) {
            s1 += 'c', s3 += 'c';
        }
        while (s1.size() < n)    s1 += 'd';
        while (s2.size() < n)    s2 += 'e';
        while (s3.size() < n)    s3 += 'f';
    }
    cout << s1 << endl;
    cout << s2 << endl;
    cout << s3 << endl;

    return 1;
}