题目描述：
Sagheer and Nubian Market
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 \(a_i\) Egyptian pounds. If Sagheer buys k items with indices x1, x2, ..., \(x_k\), then the cost of item *x**j* is \(a_{x_j}\) + \(x_j\)·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, ..., *a**n* (1 ≤ *a**i* ≤ 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 k souvenirs.
Examples
Input
Copy
3 112 3 5Output
Copy
2 11Input
Copy
4 1001 2 5 6Output
Copy
4 54Input
Copy
1 77Output
Copy
0 0Note
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.
思路：
题目是说现在有S块钱，有N个物品，物品的价格是本价加上购买的物品数乘物品编号，每个物品只能买一次。求S最多能买多少个物品，如果在买的数量相同的情况下求出用钱最少的。
刚开始，一看n是\(10^5\)量级，可以枚举来做，从n到0，每次枚举都算一下每件物品的花费，为了买到的物品最多的情况下花费要尽可能少，给每样物品的花费排一个序，再求一个花费前缀和，再用lower_bound二分查找大于S的值的位置，减一就是小于等于S的花费的位置记为pos。
下面详细说一说pos的含义：如果找到了一个位置pos（这里是前缀和，位置实际上就是购买的物品数目）花费小于等于S，这个位置pos表示了在资金足够的情况下，在枚举的物品数i的限制下，最多可以买pos个物品。如果pos等于i，就刚刚好是可以购买的最大物品数。如果pos>i，因为现在可能还没用完S，就说明有可能买更多的物品（同时买更多的物品时每个物品的花费要高些），如果pos<i，就说明现在资金用到最大程度都买不到i个物品，说明i不满足条件，继续往下枚举。（这是当时意识不太清醒时想出来的，现在来看竟然感觉不是很好理解，佩服我自己）。
由于i是从大到小枚举的，那pos>i的状态表示的购买更多物品的可能性已经枚举过了，现在只有两种可能，就是当前枚举值满足和不满足。一旦出现满足，跳出循环，输出答案。这种方法的时间复杂度是\(O(n^2log_2n)\)的，过不了太多数据。
代码：
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <cstdio>
#define max_n 100005
#define INF 0x3f3f3f3f
using namespace std;
int n;
int S;
int a[max_n];
long long sum[max_n];
template<typename T>
inline void read(T& x)
{
x=0;int f=0;char ch=getchar();
while('0'>ch||ch>'9'){if(ch=='-')f=1;ch=getchar();}
while('0'<=ch&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
x=f?-x:x;
}
#pragma optimize(2)
int main()
{
read(n);
read(S);
int minm = INF;
for(int i = 1;i<=n;i++)
{
read(a[i]);
minm = min(minm,a[i]);
}
int items = S/minm;//降一下枚举的起点
items = min(items,n);
long long money = 0;
int pos = 0;
for(int i = items;i>0;i--)
{
memset(sum,0,sizeof(sum));
for(int j = 1;j<=n;j++)
{
sum[j] = a[j] + i*j;//求花费
}
sort(sum+1,sum+n+1);
/*for(int j = 1;j<=n;j++)
{
cout << sum[j] << " ";
}
cout << endl;*/
for(int j = 1;j<=n;j++)
{
sum[j] += sum[j-1];//求花费的前缀和
}
/*for(int j = 1;j<=n;j++)
{
cout << sum[j] << " ";
}
cout << endl;
cout << endl;*/
pos = upper_bound(sum+1,sum+n+1,S)-sum-1;
//cout << "pos " << pos << endl;
if(pos>=i)
{
if(pos<=n)//这里是所有元素都小于S时的情况
{
money = sum[pos];
}
else
{
money = sum[pos-1];
}
break;
}
}
printf("%d %I64d\n",pos,money);
return 0;
}
其实枚举是对的，不过不是线性枚举，采用二分枚举，复杂度降到\(O(log_2(nlog_2n))\)。
注意返回的是记录下的最后的满足条件的枚举的物品数量。
代码：
#include <iostream>
#include <algorithm>
#define max_n 100005
using namespace std;
int n;
int S;
int a[max_n];
long long sum[max_n];
long long cost = 0;
long long mincost = 0;
int items = 0;
template<typename T>
inline void read(T& x)
{
x=0;int f=0;char ch=getchar();
while('0'>ch||ch>'9'){if(ch=='-')f=1;ch=getchar();}
while('0'<=ch&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
x=f?-x:x;
}
#pragma optimize(2)
int main()
{
read(n);
read(S);
for(int i = 1;i<=n;i++)
{
read(a[i]);
}
int l = 0;
int r = n;
int mid = 0;
while(l<=r)
{
//cout << "l " << l << " r " << r << endl;
cost = 0;
mid = (l+r)>>1;
//cout << "mid " << mid << endl;
for(int i = 1;i<=n;i++)
{
sum[i] = (long long)a[i]+(long long)mid*i;
}
sort(sum+1,sum+n+1);
for(int i = 1;i<=mid;i++)
{
cost += sum[i];
}
//cout << "cost " << cost << endl;
if(S>=cost)
{
items = mid;
mincost = cost;
l=mid+1;
}
else
{
r=mid-1;
}
}
printf("%d %I64d",items,mincost);
return 0;
}