每日一题 - 240403 - 防御力-优快云博客

本文链接：https://blog.youkuaiyun.com/CodeToLiao/article/details/137344874

防御力

TAG - $算法 - 【贪心】$
时间复杂度 - $\ast \log N)$

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

const int N = 2e6 + 7;
vector<pair<int, int> > a, b;

bool cmpA(const pair<int, int>& x, const pair<int, int>& y) {
    return x.second == y.second ? x.first < y.first : x.second < y.second;
}

bool cmpB(const pair<int, int>& x, const pair<int, int>& y) {
    return x.second == y.second ? x.first < y.first : x.second > y.second;
}

void solve() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        int ai;
        scanf("%d", &ai);
        a.push_back(make_pair(i, ai));
    }
    for (int i = 1; i <= m; i++) {
        int bi;
        scanf("%d", &bi);
        b.push_back(make_pair(i, bi));
    }
    string s;
    cin >> s;

    sort(a.begin(), a.end(), cmpA);
    sort(b.begin(), b.end(), cmpB);

    int idxA = 0, idxB = 0;
    for (int i = 0; i < s.size(); i++) {
        bool f = s[i] - '0';
        printf("%c%d\n", "AB"[f], !f ? a[idxA++].first : b[idxB++].first);
    }
    putchar('E');
}

signed main() {
    int t = 1;
    // scanf("%d", &t);
    while (t--) solve();
    return 0;
}
/*
A = 2 ^ D
B = 3 ^ D

// -- A 从小到大
A = 2 <==> [A, B, D] = [2, 3, 1]
B = 6 <==> [A, B, D] = [4, 9, 2]
A = 4 <==> [A, B, D] = [8, 27, 3]

A = 4 <==> [A, B, D] = [4, 9, 2]
B = 6 <==> [A, B, D] = [5.50, 15, 2.46 < 3]
A = 2 <==> [A, B, D] = [7.50 < 8, 24.19 < 27, 2.90 < 3]
-- //

// -- B 从大到小
B = 3 <==> [A, B, D] = [2, 3, 1]
A = 2 <==> [A, B, D] = [4, 9, 2]
B = 18 <==> [A, B, D] = [8, 27, 3]

B = 18 <==> [A, B, D] = [6.19, 18, 2.63]
A = 2 <==> [A, B, D] = [8.19, 27.90 > 27, 3.03]
B = 3 <==> [A, B, D] = [8.69 > 8, 30.90 > 27, 3.12 > 3]
-- //
*/