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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#127352 | #6632. Minimize Median | oreoioiwy# | WA | 48ms | 8060kb | C++17 | 3.3kb | 2023-07-19 16:11:03 | 2023-07-19 16:11:05 |
Judging History
answer
#include <bits/stdc++.h>
#define int long long
using namespace std;
typedef long long ll;
// #define tpyeinput int
// inline char nc() {static char buf[1000000],*p1=buf,*p2=buf;return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;}
// inline void read(tpyeinput &sum) {char ch=nc();sum=0;while(!(ch>='0'&&ch<='9')) ch=nc();while(ch>='0'&&ch<='9') sum=(sum<<3)+(sum<<1)+(ch-48),ch=nc();}
template<typename T>
void read(T &x) {
int f = 1;
x = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-')f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = x * 10 + (ch ^ 48);
ch = getchar();
}
x *= f;
}
const int N = 1e6+10;
int n, m, K;
int a[N], f[N][30];
int M[N];
void ST_pre() {
for(int i=1; i<=m; i++) f[i][0] = M[i];
int t = log(m)/ log(2) + 1;
for(int j=1; j<t; j++) {
for(int i=1; i<=m-(1<<j)+1; i++) {
f[i][j] = min(f[i][j-1], f[i+(1<<(j-1))][j-1]);
}
}
}
int query(int l, int r) {
int k = log(r-l+1)/log(2);
return min(f[l][k], f[r-(1<<k)+1][k]);
}
int ksm(int x,int y)
{
int res = 1;
while(y)
{
if(y & 1)
{
res *= x;
}
x *= x;
y >>= 1;
}
return res;
}
void precost()
{
for(int i = 4; i <= m; i++)
{
int mn = M[i];
vector<pair<int,int>> prims;
int ii = i;
for(int j = 2; j <= sqrt(i); j++)
{
if(ii % j == 0)
{
prims.push_back({j,0});
while(ii % j == 0)
{
prims.back().second++;
ii /= j;
}
}
}
if(ii > 1)
{
prims.push_back({i,1});
}
if(prims.size() == 1)
{
auto [x,y] = prims[0];
for(int j = 1; j <= (y >> 1); j++)
{
mn = min(mn, M[ksm(x,j)] + M[ksm(x,y-j)]);
}
}else
{
int sum = 0;
for(auto [x,y] : prims)
{
sum += M[ksm(x,y)];
}
mn = min(mn, sum);
}
M[i] = mn;
}
}
ll check(int mid) {
if(mid == a[n/2+1]) return 0;
if(mid > a[n/2+1]) return -1;
ll cost = 0;
int l = lower_bound(a+1, a+1+n, mid+1) - a;
// int ll = lower_bound(a+1, a+1+n, mid) - a;
int cnt = l-1;
// if(cntt > n/2) return -1;
while(cnt < n/2+1) {
int tmp = mid ? (a[l]-1)/mid+1 : a[l]+1;
cost += query(tmp, m);
cnt++; l++;
}
return cost;
}
void solve() {
read(n); read(m); read(K);
for(int i=1; i<=n; i++) read(a[i]);
for(int i=1; i<=m; i++) read(M[i]);
// M_pre();
precost();
ST_pre();
sort(a+1, a+1+n);
int l = 0, r = a[n/2+1];
while(l < r) {
int mid = l + r >> 1;
ll cost = check(mid);
if(cost >= 0 && cost <= K) r = mid;
else l = mid + 1;
}
printf("%lld\n", l);
}
signed main() {
#ifdef LOCAL_TEST
freopen("test.in", "r", stdin);
#endif
int times = 1;
read(times);
while(times--) {
solve();
}
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 2ms
memory: 7996kb
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: 48ms
memory: 8060kb
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 0 0 0 0 0 0 3 4 0 0 0 0 1 1 0 0 0 0 1 1 0 2 2 0 1 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 1 2 0 0 1 1 1 1 1 0 1 0 1 2 1 0 6 3 1 0 1 0 2 0 0 3 0 1 0 1 0 2 0 0 0 1 1 2 2 4 0 0 0 1 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 0 0 0 2 3 1 0 1 1 1 0 1 5 0 1 2 0 2 0 1 ...
result:
wrong answer 27th numbers differ - expected: '0', found: '1'