CF812C Sagheer and Nubian Market

题目描述

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 nn different items numbered from 11 to nn . The ii -th item has base cost a_{i}a**i Egyptian pounds. If Sagheer buys kk items with indices x_{1},x_{2},...,x_{k}x1,x2,...,x**k , then the cost of item x_{j}x**j is a_{xj}+x_{j}·kaxj+x**j⋅k for 1<=j<=k1<=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 kk .

Sagheer wants to buy as many souvenirs as possible without paying more than SS 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?

输入格式

The first line contains two integers nn and SS ( 1<=n<=10^{5}1<=n<=105 and 1<=S<=10^{9}1<=S<=109 ) — the number of souvenirs in the market and Sagheer's budget.

The second line contains nn space-separated integers a_{1},a_{2},...,a_{n}a1,a2,...,a**n ( 1<=a_{i}<=10^{5}1<=a**i<=105 ) — the base costs of the souvenirs.

输出格式

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

题意翻译

题目描述

Sagheer在去Luxor和 Aswan的旅途中去了Nubian市场给它的朋友和亲戚买纪念品。这个市场有一些奇怪的规则。市场上有nn件商品，标号为11到nn。第ii件商品有一个基本花费为a_{i}a**i埃及镑。如果Sagheer买了kk件下标分别为x_{1},x_{2},...,x_{k}x1,x2,...,x**k的商品，那么第jj件商品的实际花费就是a_{xj}+x_{j}·k(1 <= j <= k)axj+x**j⋅k(1<=j<=k)换句话说，每件商品的实际花费就是它的基本花费加上(它的下标×总购买件数)

Sagheer想花最多SS埃及镑来买尽量多的纪念品，注意它只能买同一种物品一件。如果有多种方式使纪念品数量最大，他会选择花费最小的方案。你能帮助他完成他的任务吗？

输入格式

第一行包含2个整数nn和SS (1 <= n <= 10^5(1<=n<=105且1 <= S <= 10^9)1<=S<=109)

第二行包含nn个以空格分开的整数a_{1},a_{2},...,a_{n}(1 <= a_{i} <= 10^5)a1,a2,...,a**n(1<=a**i<=105)

输出格式

一行，输出2个整数kk - Sagheer最大能买到的纪念品数量和TT - 最小Sagheer的总花费。

输入输出样例

输入 #1复制

输出 #1复制

输入 #2复制

输出 #2复制

输入 #3复制

输出 #3复制

说明/提示

In the first example, he cannot take the three items because they will cost him [5,9,14][5,9,14] with total cost 2828 . If he decides to take only two items, then the costs will be [4,7,11][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][5,10,17,22] .

In the third example, there is only one souvenir in the market which will cost him 88 pounds, so he cannot buy it.

题解：

2019.11.1模拟赛T1 60分场

感谢出题人给我一次重新再来的机会改到了100分。

一开始看到这道题，往DP那里想了想，后来发现明显不行，因为这道题还需要统计最多拿多少件以及拿这些件的钱数。这些信息仅用DP是维护不了的。

然后考虑贪心。贪心思路是从$n-1$枚举，反复更新当前状态下所有物品的实际价值，然后排序判断合法与否，如果合法，那么当前状态一定是最优解，直接输出$exit$就好。

时间复杂度的话应该是$O(n^2+nlogn)$（把排序算上了）过不了全部数据，只能拿60分。

后来出题人@zcs0724友情提示是二分。

所以这道题一下子就显得没难度了。我们二分的对象是最多拿多少个（这是显然的，因为最多拿多少个决定了钱数）。那么我们二分的区间就是$0-n$，注意要考虑0的情况（样例3不是特判过的）。然后$check()$函数的写法思路是和暴力思路一样的。退出二分的时候统计当前状态（最优解状态）的答案即可。

简单总结一下，我们的暴力思路是枚举判断，正解思路是二分判断，显然二分可以优化枚举的效率。这道题也是因此而把复杂度降到了log级别，然后得以AC的。所以与其把这道题说成贪心+二分，还不如说成暴力枚举的优化。（当初和七哥@littleseven讨论的时候说起了Log级别的算法，硬是没想到二分，我太菜了）

代码：

#include<cstdio>

#include<algorithm>

#define int long long

using namespace std;

const int maxn=1e5+1;

int n,s,temp,ans;

int l,r;

struct node

{

    int a,t,order;

}p[maxn];

bool cmp(node a,node b)

{

    return a.t<b.t;

}

bool check(int x)

{

    temp=0;

    for(int j=1;j<=n;j++)

        p[j].t=p[j].a+p[j].order*x;

    sort(p+1,p+n+1,cmp);

    for(int j=1;j<=x;j++)

        temp+=p[j].t;

    if(temp<=s)

        return 1;

    return 0;

}

signed main()

{

    scanf("%lld%lld",&n,&s);

    for(int i=1;i<=n;i++)

    {

        scanf("%lld",&p[i].a);

        p[i].order=i;

    }

    l=0,r=n;

    while(l<r)

    {

        int mid=(l+r+1)/2;

        if(check(mid))

            l=mid;

        else

            r=mid-1;

    }

    printf("%lld ",l);

    for(int i=1;i<=n;i++)

        p[i].t=p[i].a+p[i].order*l;

    sort(p+1,p+n+1,cmp);

    for(int i=1;i<=l;i++)

        ans+=p[i].t;

    printf("%lld",ans);

    return 0;

}