标签:space its ons http cos air push alt highlight
比赛爆零==
简单来说 题意就是 给一个N 然后给了4种操作的代价 求最小的代价。用DFS搜索
#include<bits/stdc++.h> using namespace std; typedef long long ll; int main () { ios::sync_with_stdio(false); cin.tie(0); int t; cin >> t; while(t--) { ll n, a, b, c, d; cin >> n >> a >> b >> c >> d; std::vector<pair<int, ll>> ops; // ops.emplace_back(2, a); ops.push_back(make_pair(2, a)); ops.push_back(make_pair(3, b)); ops.push_back(make_pair(5, c)); const ll inf = 4e18; map<ll, ll> mp; mp[0] = 0; mp[1] = d; function<ll(ll)> DFS = [&](ll x) { if(mp.find(x) != mp.end()) { //如果找到了返回这个值 return mp[x]; } ll ret = inf; //不进行判断会wa if((long double)d * (long double)x < ret) { ret = d * x; } for(auto& p : ops) { int num = p.first; ll cost = p.second; //分成了两种情况:第一种是减1;第二种是加1 ret = min(ret, DFS(x / num) + cost + d * (x % num)); if(x % num != 0) { ret = min(ret, DFS(x / num + 1) + cost + d * (num - x % num)); } } return mp[x] = ret; }; cout << DFS(n) << endl; } return 0; }
标签:space its ons http cos air push alt highlight
原文地址:https://www.cnblogs.com/lightac/p/12952435.html