結果

問題 No.1929 Exponential Sequence
ユーザー null
提出日時 2025-02-02 01:46:09
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 102 ms / 2,000 ms
コード長 2,771 bytes
コンパイル時間 1,799 ms
コンパイル使用メモリ 118,112 KB
実行使用メモリ 20,964 KB
最終ジャッジ日時 2025-02-02 01:46:13
合計ジャッジ時間 3,997 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge3
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 4
other AC * 24
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <iostream>
#include <vector>
#include <functional>
#include <algorithm>
using namespace std;
typedef long long ll;
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    ll S;
    cin >> n >> S;
    vector<ll> a(n);
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    
    // For each index, find the maximum exponent k (starting at 1) such that a[i]^k <= S.
    // For each index i, possible exponents are 1,2,...,maxExp[i].
    vector<int> maxExp(n, 0);
    for (int i = 0; i < n; i++){
        ll power = 1;
        int cnt = 0;
        // We must take at least k=1.
        while (true) {
            // Check for overflow: if power > S/a[i], then multiplying by a[i] would exceed S.
            if (power > S / a[i]) break;
            power *= a[i];
            cnt++;
        }
        if(cnt < 1){ 
            // If even a[i]^1 > S, no valid sequence exists.
            cout << 0 << "\n";
            return 0;
        }
        maxExp[i] = cnt;
    }
    
    // Split indices into two groups.
    int mid = n / 2;
    vector<int> groupAIndices, groupBIndices;
    for (int i = 0; i < mid; i++){
        groupAIndices.push_back(i);
    }
    for (int i = mid; i < n; i++){
        groupBIndices.push_back(i);
    }
    
    // For a given group, recursively enumerate all possible sums.
    auto recGroup = [&](auto &recGroup, int pos, ll sum, const vector<int>& indices, vector<ll>& store) -> void {
        if (pos == indices.size()){
            store.push_back(sum);
            return;
        }
        int i = indices[pos];
        ll power = 1;
        // Try exponents from 1 up to maxExp[i] (if the sum does not exceed S).
        for (int exp = 1; exp <= maxExp[i]; exp++){
            if(exp == 1)
                power = a[i];  // a[i]^1
            else
                power *= a[i]; // a[i]^exp, computed iteratively.
            if(sum + power > S) break;  // further exponents only increase the sum
            recGroup(recGroup, pos + 1, sum + power, indices, store);
        }
    };
    
    vector<ll> groupASums, groupBSums;
    recGroup(recGroup, 0, 0LL, groupAIndices, groupASums);
    recGroup(recGroup, 0, 0LL, groupBIndices, groupBSums);
    
    // Sort one of the groups (here groupBSums) for binary search.
    sort(groupBSums.begin(), groupBSums.end());
    
    // For each sum from group A, count how many sums from group B satisfy: sum_A + sum_B <= S.
    ll answer = 0;
    for (ll s : groupASums){
        ll remain = S - s;
        // upper_bound finds the first element greater than remain.
        auto it = upper_bound(groupBSums.begin(), groupBSums.end(), remain);
        answer += (it - groupBSums.begin());
    }
    
    cout << answer << "\n";
    return 0;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0