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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#734215 | #6632. Minimize Median | Vedant18 | WA | 125ms | 3852kb | C++23 | 3.4kb | 2024-11-11 06:40:34 | 2024-11-11 06:40:34 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define in(a, b) for (ll i = (a); i <= (b); i++) // in using i
#define inj(a, b) for (ll j = (a); j <= (b); j++) // in using j
#define ink(a, b) for (ll k = (a); k <= (b); k++) // in using k
#define inl(a, b) for (ll l = (a); l <= (b); l++) // in using l
#define inr(a, b) for(ll i = (a); i >= (b); i--) // in reverse
#define inrj(a, b) for(ll j = (a); j >= (b); j--) // in reverse
#define inrk(a, b) for(ll k = (a); k >= (b); k--) // in reverse
#define inrl(a, b) for(ll l = (a); l >= (b); l--) // in reverse
#define tt ll tcs; cin>>tcs; while(tcs--) // include test cases
#define ina(arr,n) ll arr[(n+1)]={0}; in(1,n) cin>>arr[i] // input arr of n elements
#define inv(vec,n) vector<ll> vec(n+1); vec[0]=0; in(1,n) cin>>vec[i]; // input vector of n elements
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define pll pair <ll,ll>
#define vpll vector <pll>
#define sll set <ll>
#define vll vector<ll>
#define mll map <ll,ll>
#define all(x) x.begin(), x.end()
const ll mod=1e9+7;
#define vvll vector<vll>
#define pref(p,a,n) vll p(n+1); in(1,n) p[i]=p[i-1]+a[i];
#define vec2(a,n,m) vvll a(n+1,vll(m+1))
// #define vec2(a,n,m,val) vvll a(n+1,vll(m+1,val))
#define vec3(a,l,m,n) vector<vvll>a(l+1,vvll(m+1,vll(n+1)));
// #define vec3(a,l,m,n,val) vector<vvll>a(l+1,vvll(m+1,vll(n+1,val)));
# define FAST ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
int main(){
FAST
tt{
ll n,m,k;
cin>>n>>m>>k;
inv(a,n);
inv(cst,m);
//O(m)
inr(m-1,1){
cst[i]=min(cst[i],cst[i+1]);
}
cst.push_back(1e17);
cst[1]=0;
unordered_multiset<ll>future;
future.insert(1e17);
vvll vals(m+2);
for(ll i=2;i<=(m+1);i++){
//O(mlogm) harmonic
if(i==(m+1)){
int t=1;
}
cst[i]=min(cst[i],*future.begin());
for(ll j=2*i;j<=i*i;j+=i){
ll nv=cst[i]+cst[j/i];
if(j>m){
future.insert(nv);
break;
}
vals[j].push_back(nv);
future.insert(nv);
}
for(auto x:vals[i]){
future.erase(future.find(x));
}
// cout<<"DONE"<<i<<endl;
}
// in(1,m){
// cout<<cst[i]<<' ';
// }
// cout<<endl;
ll lo=0;
ll hi=m;
auto ok=[&](ll md)->bool{
//can i make more than equal to (n)/2 elements less than equal to md
vll cc;
in(1,n){
// ll x=(a[i]+md-1)/md;
ll x=(a[i]/(md+1))+1;
cc.push_back(cst[x]);
}
sort(all(cc));
ll tcst=0;
for(int i=0;i<((n+1)/2);i++){
tcst+=cc[i];
}
return (tcst<=k);
}
;
// ok(3);
ll ans=m;
while(lo<=hi){
ll mid=(lo+hi)/2;
if(ok(mid)){
ans=mid;hi=mid-1;
}
else{
lo=mid+1;
}
}
cout<<ans<<endl;
}
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3852kb
input:
3 3 5 0 2 5 2 3 2 4 6 13 3 5 3 2 5 3 3 2 4 6 13 3 5 6 2 5 2 3 2 4 6 13
output:
2 2 1
result:
ok 3 number(s): "2 2 1"
Test #2:
score: -100
Wrong Answer
time: 125ms
memory: 3628kb
input:
100000 5 10 5 3 7 1 10 10 11 6 11 6 1 8 9 1 3 1 5 6 51 2 2 2 5 1 42 61 26 59 100 54 5 10 76 7 5 8 4 7 97 4 44 83 61 45 24 88 44 44 5 8 90 1 1 5 1 3 35 15 53 97 71 83 26 7 5 3 52 1 1 3 1 1 22 6 93 5 6 28 6 6 1 3 1 9 31 2 19 10 27 5 8 31 3 6 2 1 2 32 29 13 7 57 34 9 5 5 6 75 3 3 4 5 4 40 56 38 60 17 3...
output:
0 2 1 0 0 0 0 0 3 4 0 0 0 0 1 1 0 0 0 0 1 1 0 2 2 0 0 0 0 0 2 0 0 0 2 2 0 1 0 0 0 0 1 0 2 4 1 1 0 0 2 0 0 7 0 1 0 0 0 1 1 0 1 0 1 0 1 2 1 0 6 3 0 0 1 0 2 0 0 3 0 1 0 1 0 2 0 0 0 1 1 2 1 4 0 0 0 0 0 0 1 2 2 1 2 2 0 1 1 0 0 0 0 0 1 2 1 4 1 0 4 1 2 1 0 0 0 0 1 2 1 0 0 2 3 1 0 1 1 1 0 1 5 0 1 2 0 2 1 1 ...
result:
wrong answer 3rd numbers differ - expected: '0', found: '1'