D. Gadgets for dollars and pounds

原题链接

题解

1.如果限定在 $x$ 天内买完，那么限定在 $x+1$ 天内也能买完，二分浮现。
2.如果要买 $type1$ ，那么一定是在这 $x$ 天内 $a$ 价格最低的那天一次买齐，且优先买价格低的 $type1$
3.优先买 单价x汇率 最低的那个

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll N = 114514;
ll n, m, k, s;

ll a[2 * N], b[2 * N];

struct unit {
    ll id, val;
    bool operator < (const unit &other) const {
        return val < other.val;
    }
};

vector<unit> dollar, pounds;

struct node {
    ll id, day;
} ans[2 * N];

struct {
    ll day, val;
} prea[2 * N], preb[2 * N];

bool check(ll x) {
    ll it1 = 0, it2 = 0;
    ll sum = 0;

    for (ll i = 1; i <= k; i++) {
        if (it1 < dollar.size() && it2 < pounds.size()) {
            if (dollar[it1].val * prea[x].val < pounds[it2].val * preb[x].val) {
                sum += dollar[it1].val * prea[x].val;
                it1++;
            } else {
                sum += pounds[it2].val * preb[x].val;
                it2++;
            }
        } else if (it1 < dollar.size()) {
            sum += dollar[it1].val * prea[x].val;
            it1++;
        } else {
            sum += pounds[it2].val * preb[x].val;
            it2++;
        }
    }

    return sum <= s;
}

void build(ll x) {
    ll it1 = 0, it2 = 0;

    for (ll i = 1; i <= k; i++) {
        if (it1 < dollar.size() && it2 < pounds.size()) {
            if (dollar[it1].val * prea[x].val < pounds[it2].val * preb[x].val) {
                ans[i].id = dollar[it1].id;
                ans[i].day = prea[x].day;
                it1++;
            } else {
                ans[i].id = pounds[it2].id;
                ans[i].day = preb[x].day;
                it2++;
            }
        } else if (it1 < dollar.size()) {
            ans[i].id = dollar[it1].id;
            ans[i].day = prea[x].day;
            it1++;
        } else {
            ans[i].id = pounds[it2].id;
            ans[i].day = preb[x].day;
            it2++;
        }
    }
}

void solve() {
    cin >> n >> m >> k >> s;

    for (ll i = 1; i <= n; i++) cin >> a[i];
    for (ll i = 1; i <= n; i++) cin >> b[i];

    for (ll i = 1; i <= m; i++) {
        ll type, val;
        cin >> type >> val;
        if (type == 1) dollar.push_back({i, val});
        else pounds.push_back({i, val});
    }

    sort(dollar.begin(), dollar.end());
    sort(pounds.begin(), pounds.end());

    prea[0].val = 2e9;
    prea[0].day = -1;
    preb[0].val = 2e9;
    preb[0].day = -1;

    for (ll i = 1; i <= n; i++) {
        prea[i].val = prea[i - 1].val;
        prea[i].day = prea[i - 1].day;
        preb[i].val = preb[i - 1].val;
        preb[i].day = preb[i - 1].day;

        if (a[i] < prea[i - 1].val) {
            prea[i].val = a[i];
            prea[i].day = i;
        }

        if (b[i] < preb[i - 1].val) {
            preb[i].val = b[i];
            preb[i].day = i;
        }
    }

    ll l = 0, r = n + 1;
    while (l + 1 < r) {
        ll mid = (l + r) / 2;
        if (check(mid)) r = mid;
        else l = mid;
    }

    if (r == n + 1) cout << "-1\n";
    else {
        cout << r << '\n';
        build(r);
        for (ll i = 1; i <= k; i++) cout << ans[i].id << ' ' << ans[i].day << '\n';
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ll t = 1;
    while (t--) solve();
    return 0;
}