Submission #1276470

#	Time	Username	Problem	Language	Result	Execution time	Memory
1276470		mehrzad	축제 (IOI25_festival)	C++17	12 / 100	1036 ms	2162688 KiB

festival

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

using i128 = __int128_t;

struct Coupon {
    int idx;
    long long P;
    int T;
};

static bool cmpMul(const Coupon& a, const Coupon& b) {
    // Compare a.P * a.T / (a.T-1)  <  b.P * b.T / (b.T-1)
    // cross multiply to avoid division:
    // a.P * a.T * (b.T-1)  <  b.P * b.T * (a.T-1)
    i128 left  = (i128)a.P * (i128)a.T * (i128)(b.T - 1);
    i128 right = (i128)b.P * (i128)b.T * (i128)(a.T - 1);
    return left < right;
}

static bool cmpOne(const Coupon& a, const Coupon& b) {
    return a.P < b.P;
}

std::vector<int> max_coupons(int A, std::vector<int> P, std::vector<int> T) {
    int N = (int)P.size();
    vector<Coupon> mul, one;
    mul.reserve(N);
    one.reserve(N);
    for (int i = 0; i < N; ++i) {
        Coupon c{ i, (long long)P[i], T[i] };
        if (T[i] == 1) one.push_back(c);
        else mul.push_back(c);
    }

    // sort multiplier coupons by ascending C = P*T/(T-1)
    sort(mul.begin(), mul.end(), cmpMul);
    // sort T=1 coupons by price ascending
    sort(one.begin(), one.end(), cmpOne);

    int M = (int)mul.size();
    const i128 INF = (i128)1 << 120;   // huge sentinel

    // DP[i][k] = max tokens after considering first i mul-coupons and taking exactly k of them
    vector<vector<i128>> dp(M + 1, vector<i128>(M + 1, -1));
    vector<vector<char>> took(M + 1, vector<char>(M + 1, 0));
    dp[0][0] = (i128)A;

    for (int i = 1; i <= M; ++i) {
        const Coupon& c = mul[i - 1];
        for (int k = 0; k <= i; ++k) {
            // skip case
            if (dp[i - 1][k] != -1) {
                dp[i][k] = dp[i - 1][k];
                took[i][k] = 0;
            }
            // take case
            if (k > 0 && dp[i - 1][k - 1] != -1) {
                i128 prev = dp[i - 1][k - 1];
                if (prev >= (i128)c.P) {
                    i128 after = (prev - (i128)c.P) * (i128)c.T;
                    if (after > INF) after = INF;
                    if (after > dp[i][k]) {
                        dp[i][k] = after;
                        took[i][k] = 1;
                    }
                }
            }
        }
    }

    // prefix sums for T=1 coupons
    int N1 = (int)one.size();
    vector<i128> pref(N1 + 1, 0);
    for (int i = 1; i <= N1; ++i) {
        pref[i] = pref[i - 1] + (i128)one[i - 1].P;
    }

    int bestK = 0, bestOneCnt = 0, bestTotal = 0;
    for (int k = 0; k <= M; ++k) {
        if (dp[M][k] == -1) continue;
        i128 tokens = dp[M][k];
        // find max number of T=1 coupons we can afford
        int lo = 0, hi = N1;
        while (lo < hi) {
            int mid = (lo + hi + 1) >> 1;
            if (pref[mid] <= tokens) lo = mid;
            else hi = mid - 1;
        }
        int cntOne = lo;
        int total = k + cntOne;
        if (total > bestTotal) {
            bestTotal = total;
            bestK = k;
            bestOneCnt = cntOne;
        }
    }

    // reconstruct chosen multiplier coupons
    vector<int> orderMul;
    int i = M, k = bestK;
    while (i > 0) {
        if (took[i][k]) {
            orderMul.push_back(mul[i - 1].idx);
            --k;
        }
        --i;
    }
    reverse(orderMul.begin(), orderMul.end());

    // take first bestOneCnt cheap T=1 coupons
    vector<int> orderOne;
    for (int idx = 0; idx < bestOneCnt; ++idx) {
        orderOne.push_back(one[idx].idx);
    }

    // concatenate results
    vector<int> result;
    result.reserve(bestTotal);
    result.insert(result.end(), orderMul.begin(), orderMul.end());
    result.insert(result.end(), orderOne.begin(), orderOne.end());

    return result;
}

• Subtask 0: 0 / 0
• Subtask 1: 5 / 5
• Subtask 2: 7 / 7
• Subtask 3: 0 / 12
• Subtask 4: 0 / 15
• Subtask 5: 0 / 27
• Subtask 6: 0 / 16
• Subtask 7: 0 / 18