版权声明:博主的博客不值钱随便转载但要注明出处 https://blog.csdn.net/easylovecsdn/article/details/85004075
题意:
给你n个瓶子,每个瓶子里有vi升饮料,现在又给你一个桶,让你在尽可能让装有最少饮料瓶子中的饮料多的情况下装满桶,若不能输出-1.
思路:
先选出装有最少饮料的瓶子,先让其他瓶子去填装桶,且其他瓶子填装时所剩余饮料不能少于最少瓶子中的升数,若其他瓶子能将桶填满,输出最小瓶子升数,若不能(此时所有瓶子饮料数相同)再一同填补。
Code:
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1005;
LL s;
int n, a[maxn];
int main()
{
cin >> n >> s;
//n = 1000;
//s = 980103855476;
LL sum = 0;
for (int i = 0; i < n; i++)
{
cin >> a[i];
//a[i] = 1000000000;
if (sum <= s + 1) sum += a[i];
}
if (sum == s) cout << 0 << endl;
else if (sum < s) cout << -1 << endl;
else
{
sort(a, a + n);
LL pour = 0;
for (int i = 1; i < n; i++)
{
pour += (a[i] - a[0]);
if (pour >= s) break;
}
if (pour >= s) cout << a[0] << endl;
else
{
//cout << pour << endl;
LL r = s - pour;
//cout << r << endl;
int t = r % n == 0 ? (r / n) : (r / n + 1);
a[0] -= t;
if (a[0] <= 0) cout << 0 << endl;
else cout << a[0] << endl;
}
}
return 0;
}