coderforce Educational Codeforces Round 44 (Rated for Div. 2) C(赛后补题)
2 seconds
256 megabytes
standard input
standard output
You have m = n·k wooden staves. The i-th stave has length ai. You have to assemble n barrels consisting of k staves each, you can use any k staves to construct a barrel. Each stave must belong to exactly one barrel.
Let volume vj of barrel j be equal to the length of the minimal stave in it.
You want to assemble exactly n barrels with the maximal total sum of volumes. But you have to make them equal enough, so a difference between volumes of any pair of the resulting barrels must not exceed l, i.e. |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.
Print maximal total sum of volumes of equal enough barrels or 0 if it's impossible to satisfy the condition above.
The first line contains three space-separated integers n, k and l (1 ≤ n, k ≤ 105, 1 ≤ n·k ≤ 105, 0 ≤ l ≤ 109).
The second line contains m = n·k space-separated integers a1, a2, ..., am (1 ≤ ai ≤ 109) — lengths of staves.
Print single integer — maximal total sum of the volumes of barrels or 0 if it's impossible to construct exactly n barrels satisfying the condition |vx - vy| ≤ l for any 1 ≤ x ≤ n and 1 ≤ y ≤ n.
4 2 12 2 1 2 3 2 2 3
7
2 1 010 10
20
1 2 15 2
2
3 2 11 2 3 4 5 6
0
Note
In the first example you can form the following barrels: [1, 2], [2, 2], [2, 3], [2, 3].
In the second example you can form the following barrels: [10], [10].
In the third example you can form the following barrels: [2, 5].
In the fourth example difference between volumes of barrels in any partition is at least 2 so it is impossible to make barrels equal enough.
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<stack>
using namespace std;
typedef long long LL;
LL arr[100010];
LL tem[100010];int main()
{LL sum=0;LL n,k;LL l;cin>>n>>k>>l;for(LL i=1;i<=n*k;i++){cin>>arr[i];}sort(arr+1,arr+1+n*k);LL j=0;tem[++j]=1;int ans;for(LL i=1;i<=n*k;i++){if(arr[i+1]!=arr[i])tem[++j]=i;else tem[j]=i;if(arr[i]==l+arr[1])ans=i;else if(arr[i]<l+arr[1]&&l+arr[1]<arr[i+1])ans=i;}LL pre=1;if(ans<n){cout<<"0";return 0;}if(n==1){cout<<arr[1];return 0;}for(LL m=1;m<=n;m++){if(pre+k-1+(n-m)<=ans&&k!=0){sum+=arr[pre];pre=pre+k;}else{while(--k){if(pre+k-1+(n-m)==ans){sum+=arr[pre];pre=pre+k;k=0;break;}}}if(k==0){for(LL i=pre;i<=ans;i++){sum+=arr[i];}break;}}cout<<sum;return 0;
} coderforce Educational Codeforces Round 44 (Rated for Div. 2) C(赛后补题)相关推荐
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