2018-BNUZ-ACM-GDCPC选拔赛 CodeForces 812C Sagheer and Nubian Market【补】

On his trip to Luxor and Aswan, Sagheer went to a Nubian market to buy some souvenirs for his friends and relatives. The market has some strange rules. It contains n different items numbered from 1 to n. The i-th item has base cost ai Egyptian pounds. If Sagheer buys kitems with indices x1, x2, ..., xk, then the cost of item xj is axj + xj·k for 1 ≤ j ≤ k. In other words, the cost of an item is equal to its base cost in addition to its index multiplied by the factor k.

Sagheer wants to buy as many souvenirs as possible without paying more than S Egyptian pounds. Note that he cannot buy a souvenir more than once. If there are many ways to maximize the number of souvenirs, he will choose the way that will minimize the total cost. Can you help him with this task?

Input

The first line contains two integers n and S (1 ≤ n ≤ 105 and 1 ≤ S ≤ 109) — the number of souvenirs in the market and Sagheer's budget.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the base costs of the souvenirs.

Output

On a single line, print two integers k, T — the maximum number of souvenirs Sagheer can buy and the minimum total cost to buy these ksouvenirs.

    Examples 
  
      input 
     
       Copy 
     
3 11
2 3 5

      output 
     
       Copy 
     
2 11

      input 
     
       Copy 
     
4 100
1 2 5 6

      output 
     
       Copy 
     
4 54

      input 
     
       Copy 
     
1 7
7

      output 
     
       Copy 
     
0 0

Note

In the first example, he cannot take the three items because they will cost him [5, 9, 14] with total cost 28. If he decides to take only two items, then the costs will be [4, 7, 11]. So he can afford the first and second items.

In the second example, he can buy all items as they will cost him [5, 10, 17, 22].

In the third example, there is only one souvenir in the market which will cost him 8 pounds, so he cannot buy it.

题意：有个奇怪的商店，其中的有n个物品，基础为价格ai，预算为s，每个商品的价格是会实时变动的， price[i] = a[i] + i * k 【k为所选物品的数量】，求花最少的钱买最多的物品。输出 k 和总花费

题解：对k进行二分，将算好价格的数组排序（这里我用了优先队列）后算出总价格，判断总价格小于预算s 就好了

AC代码：

#include <bits/stdc++.h>
#define ll long long

using namespace std;

int main() {
	ll n, s, a[100010];
	cin >> n >> s;
	for (ll i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
	}
	ll r = n, l = 1, mid, sum, ans1 = 0, ans2 = 0;
	while (l <= r) {
		mid = (l + r) >> 1;
//cout << mid << "--" << endl;
		priority_queue<ll, vector<ll>, greater<ll> > q;
		for (int i = 1; i <= n; i++) {
			q.push(a[i] + i*mid);
		}
		sum = 0;
		for (int i = 1; i <= mid; i++) {
			sum += q.top();
			q.pop();
		}
//cout <<"sum=" << sum << endl;
		if (sum > s) {
			r = mid - 1;
		} else {
			l = mid + 1;
			ans1 = mid;
			ans2 = sum;
		}
//cout << "ans1=" << ans1 << " ans2=" << ans2 << endl;
	}
	cout << ans1 << " " << ans2 << endl;
}

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后记。。

被小Q吐槽说这种排序方式不太健康，浪费内存，还有二分最好写成封装。

附上小Q的代码：

#include <bits/stdc++.h>

#define MAXN 100005
#define ll long long
#define pll pair<ll, ll>

using namespace std;

ll a[MAXN];
pll b[MAXN];
ll n;
ll s;

ll getval(ll p) {
    for (ll i = 1; i <= n; i++) {
        b[i].second = i;
        b[i].first = p * i + a[i];
    }
    sort(b + 1, b + n + 1);
    ll ans = 0;
    for (ll i = 1; i <= p; i++) {
        ans += b[i].first;
    }
    return ans;
}

ll binary_search(ll l, ll r) {
    ll mid = l + r >> 1;
    ll ans = mid;
    while (l <= r) {
        mid = l + r >> 1;
        if (s >= getval(mid)) {
            ans = mid;
            l = mid + 1;
        } else {
            r = mid - 1;
        }
    }
    return ans;
}

int main() {
    cin >> n >> s;
    for (ll i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
    }
    ll p = binary_search(0, n);
    printf("%lld %lld\n", p, getval(p));
}

2018-BNUZ-ACM-GDCPC选拔赛 CodeForces 812C Sagheer and Nubian Market【补】

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