Educational Codeforces Round 3
Educational Codeforces Round 3
https://codeforces.com/contest/609
3/6：ABC
D赛后过了
CD题解等下写
A. USB Flash Drives
前缀和
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int n, a[N];
int main () {
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i++) cin >> a[i];
sort (a + 1, a + n + 1, greater<int>());
for (int i = 1; i <= n; i++) {
m -= a[i];
if (m <= 0) {
cout << i;
break;
}
}
}
B. The Best Gift
容斥思想？
任意 \(n\) 个数中选两个的方案数是 \(C_{n}^{2}\)
其中除去各相同的方案数就是答案。
#include <bits/stdc++.h>
#define int long long
using namespace std;
int a[15];
signed main () {
int n, m;
cin >> n >> m;
for (int i = 1; i <= n; i++) {
int x;
cin >> x;
a[x] ++;
}
int ans = n * (n - 1) / 2;
for (int i = 1; i <= m; i++) {
ans -= a[i] * (a[i] - 1) / 2;
}
cout << ans;
}
C. Load Balancing
小思维，考虑到总和是不变的。所以最终所有数字的局面是固定的，那么只需排序之后计算差值即可。
最终局面就是：
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e5 + 5;
int a[N], sum, ans, n;
signed main () {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i], sum += a[i];
sort (a + 1, a + n + 1);
//cout << sum << endl;
int val = sum / n, cnt = sum - val * n;
//cout << val << ' ' << cnt << endl; //有n - cnt个val, cnt个val + 1
for (int i = 1; i <= n - cnt; i++) {
ans += abs (a[i] - val);
}
for (int i = n - cnt + 1; i <= n; i++) {
ans += abs (a[i] - (val + 1));
}
cout << ans / 2 << endl;
}
D. Gadgets for dollars and pounds
二分。
// LUOGU_RID: 97555092
#include <bits/stdc++.h>
#define ll long long
using namespace std;
typedef pair<int, int> pii;
const int N = 2e5 + 5, inf = 1e9;
int n, m, k, s;
pair<ll, int> a[N], b[N];
int ss[3], id[3];
pii result[N];
struct Node {
int id, type;
ll cost;
}p[N];
bool cmp (Node p1, Node p2) {
ll sum1 = 0, sum2 = 0;
return 1ll * ss[p1.type] * p1.cost < 1ll * ss[p2.type] * p2.cost;
}
bool check (int mid) {
id[1] = a[mid].second, id[2] = b[mid].second;
ss[1] = a[mid].first, ss[2] = b[mid].first;
sort (p + 1, p + m + 1, cmp);
//for (int i = 1; i <= m; i++) cout << ss[p[i].type] * p[i].cost << ' ';
//cout << endl;
ll sum = 0;
for (int i = 1; i <= k; i++) {
sum += 1ll * ss[p[i].type] * p[i].cost;
//cout << sum << ' ';
if (sum > s) return false;
}
for (int i = 1; i <= k; i++) {
result[i].first = p[i].id;
result[i].second = id[p[i].type];
}
return true;
}
signed main () {
cin >> n >> m >> k >> s;
ll mina = 1e9, minb = 1e9;
int id1, id2;
for (int i = 1; i <= n; i++) {
ll x; cin >> x;
if (x < mina) mina = x, id1 = i;
a[i] = {mina, id1};
}
for (int i = 1; i <= n; i++) {
ll x; cin >> x;
if (x < minb) minb = x, id2 = i;
b[i] = {minb, id2};
}
for (int i = 1; i <= m; i++) {
int x;
ll y;
cin >> x >> y;
p[i] = {i, x, y};
}
int l = 1, r = n, ans = -1;
while (l <= r) {
int mid = l + r >> 1;
if (check (mid)) {
ans = mid, r = mid - 1;
}
else l = mid + 1;
//cout << l << ' ' << r << endl;
}
if (ans == -1) cout << "-1\n";
else {
cout << ans << endl;
for (int i = 1; i <= k; i++) cout << result[i].first << ' ' << result[i].second << endl;
}
}
//n天,m物品,买k个,现有s元
//两种货币a,b
//给出n天对应的汇率以及所有物品的两种价格
//最快能在第几天把k件物品买完,在哪天买的第几种物品
//一种商品只能购买一次，但是一天可以买多种商品。
//钱是可以一次性转换的
