E. Klee's SUPER DUPER LARGE Array!!!
https://codeforces.com/contest/2009/problem/E
思路：
题目让我们求从k开始的n个数的前k个数的和与剩下的数的和的差最小是多少，可以用数学思维O(1)求解，都是我数学比较差，我们这里用二分。
代码：
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<algorithm>
#define int long long
#define pb push_back
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define lowbit(x) x&(-x)
#define pll pair<int,int>
const int N = 3e5 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9+7;
using namespace std;
int ss(int a, int b) {
return ((a + b) * (b - a + 1)) / 2;
}
void solve() {
int n, k;
cin >> n >> k;
int l = 1, r = n-1;
int ans;
while (l <= r) {
int mid = l + r >> 1;
if (ss(k, k + mid - 1) < ss(k + mid, k + n - 1)) ans=mid,l = mid + 1;
else ans=mid,r = mid - 1;
}
int res = 1e20;
for (int i = max(1ll,ans - 10); i <= ans + 10&&i<n; i++) {
res = min(res, abs(ss(k, k + i - 1) - ss(k + i, k + n - 1)));
}
cout << res << '\n';
}
signed main() {
ios; TEST
solve();
return 0;
}D. Wooden Toy Festival
https://codeforces.com/contest/1840/problem/D
思路：
很明显是一道二分答案的题目，所以我们只需要考虑判断函数怎么写即可，我们发现我们只需要安排的雕刻师不大于三即可，我们可以对数组排序，我们看看每一个雕刻师最多保证答案不会超出界限，最后判断我们使用的雕刻师数量即可。
代码：
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<algorithm>
#define int long long
#define pb push_back
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define lowbit(x) x&(-x)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
int n, a[N];
bool check(int x) {
//安排几个人
int cnt = 0;
int pre = a[1];
for (int i = 1; i <= n; i++) {
if ((a[i] - pre + 1) / 2 <= x) {
}
else {
pre = a[i];
cnt++;
}
if (cnt >= 3) return false;
}
return cnt < 3;
}
void solve()
{
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + 1 + n);
int l = 0, r = 1e9;
int ans;
while (l <= r) {
int mid = l + r >> 1;
if (check(mid)) ans=mid,r = mid-1;
else l = mid + 1;
}
cout << ans << '\n';
}
signed main() {
ios; TEST
solve();
return 0;
}
分巧克力
1227. 分巧克力 - AcWing题库
思路：
只要知道一个长为h，宽为w的巧克力可以分成（h/x)*(w/x)个边长为x的正方形巧克力即可。
代码：
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<algorithm>
#define int long long
#define pb push_back
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define lowbit(x) x&(-x)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
int n, k;
int h[N], w[N];
bool check(int x) {
int res = 0;
for (int i = 1; i <= n; i++) {
res += (h[i] / x) * (w[i] / x);
if (res >= k) return true;
}
return res >= k;
}
void solve()
{
cin >> n >> k;
for (int i = 1; i <= n; i++) cin >> h[i] >> w[i];
int l = 1, r = 1e9, ans;
while (l <= r) {
int mid = l + r >> 1;
if (check(mid)) {
ans = mid;
l = mid + 1;
}
else {
r = mid - 1;
}
}
cout << ans << '\n';
}
signed main() {
ios;
solve();
return 0;
}
4001. 训练
4001. 训练 - AcWing题库
思路：
用二分求出每个位置有多少个数比他小，再在每次的关系中判断，如果有关系的位置中，恰好这个位置是比你小的数，则答案减一。
代码：
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<algorithm>
#define int long long
#define pb push_back
#define TEST int T; cin >> T; while (T--)
#define ios ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr)
#define lowbit(x) x&(-x)
#define pll pair<int,int>
const int N = 1e6 + 30;
const int M = 1e3 + 10;
const int inf = 512785182741247112;
const int mod = 1e9 + 7;
using namespace std;
bool cmp(pll a, pll b) {
return a.first < b.first;
}
void solve()
{
int n, k;
cin >> n >> k;
vector<pll>r(n + 1);
vector<int>a(n + 1);
vector<int>b(n + 1);
for (int i = 1; i <= n; i++) {
cin >> r[i].first;
r[i].second = i;
a[i] = r[i].first;
b[i] = r[i].first;
}
vector<int>cnt(n + 1);
sort(r.begin() + 1, r.end(), cmp);
sort(a.begin() + 1, a.end());
for (int i = 1; i <= n; i++) {
int pos=lower_bound(a.begin()+1, a.end(), a[i]) - a.begin();
cnt[r[i].second] = pos - 1;
}
for (int i = 1; i <= k; i++) {
int x, y;
cin >> x >> y;
if (b[x] > b[y]) cnt[x] = max(0ll, cnt[x] - 1);
if (b[y] > b[x]) cnt[y] = max(0ll, cnt[y] - 1);
}
for (int i = 1; i <= n; i++) cout << cnt[i] << ' ';
}
signed main() {
ios;
solve();
return 0;
}
