【AtCoder ABC 127】补题AK E待补
A 语法 模拟
思路
几个else if 就行代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "Yes" << endl
#define cn cout << "No" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
int a, b;
int ans;
int main() {
cin >> a >> b;
if (a >= 13) ans = b;
else if (a >= 6 && a <= 12)
ans = b / 2;
else
ans = 0;
cout << ans;
return 0;
}
B 语法 模拟
思路
按题意写就行代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "Yes" << endl
#define cn cout << "No" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
ll r, d, x;
int main() {
cin >> r >> d >> x;
forn(i, 1, 10) x = x * r - d, cout << x << endl;
return 0;
}
C 思维
思路
第i次，暴力将l 到r 加上1（这些ID卡可通过这扇门）。最后再看每一张ID卡，是否m条门都打得开。但这很明显会TLE飞。 很明显区间修改，直接上差分优化就行了代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "Yes" << endl
#define cn cout << "No" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 1e5 + 5;
int n, m;
int l, r;
int d[N];
int main() {
cin >> n >> m;
forn(i, 1, m) {
cin >> l >> r;
d[l]++, d[r + 1]--;
}
int ans = 0;
forn(i, 1, n) d[i] += d[i - 1];
forn(i, 1, n) if (d[i] == m) ans++;
cout << ans;
return 0;
}
D 贪心
思路
排个序，每次把尽可能小的改成尽可能大的，如果改不了就break代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "Yes" << endl
#define cn cout << "No" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
ll n, m;
struct node {
ll x, y;
} s[N];
bool cmp(node a, node b) { return a.y > b.y; }
ll a[N];
ll ans;
int main() {
cin >> n >> m;
forn(i, 1, n) cin >> a[i];
forn(i, 1, m) cin >> s[i].x >> s[i].y;
sort(a + 1, a + n + 1);
sort(s + 1, s + m + 1, cmp);
int cnt = 1;
forn(i, 1, m) {
if (cnt > n) break;//没加这个就RE了，我猜是爆int成负数了
forn(j, cnt, min(n, cnt + s[i].x - 1)) {
if (a[j] > s[i].y) break;
a[j] = s[i].y;
}
cnt += s[i].x;
}
forn(i, 1, n) ans += a[i];
cout << ans;
return 0;
}
E 排列组合
思路
不会，抄题解结论的代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "Yes" << endl
#define cn cout << "No" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
ll p = 1e9 + 7;
ll qpow(ll a, ll b) {
ll s = 1;
while (b) {
if (b & 1) s = (s * a) % p;
a = (a * a) % p;
b >>= 1;
}
return s;
}
ll C(ll a, ll b) {
ll res = 1;
forn_(i, a, a - b + 1) res = (res * i) % p;
ll fac = 1;
forn(i, 1, b) fac = (fac * i) % p;
res = (res * qpow(fac, p - 2)) % p;
return res;
}
ll n, m, k;
ll ans;
int main() {
cin >> n >> m >> k;
forn(i, 1, n) ans = (ans + (n - i) * i * m * m) % p;
forn(i, 1, m) ans = (ans + (m - i) * i * n * n) % p;
ans = (ans * C(n * m - 2, k - 2)) % p;
cout << ans;
return 0;
}
F 数学 数据结构
思路
观察下原式，就是几个绝对值加一起，根据绝对值的几何关系，选中位数能让最小。那么问题就转化为怎么求中位数了，直接上对顶堆
还有个问题就是怎么求和，如果遍历一个一个加的话，肯定TLE了，考虑优化。
因为对顶堆的性质，一个堆全大于中位数，一个堆全小于中位数，那么就很好去绝对值了，用个值s来维护堆里所有元素的和，最后直接算中位数乘以堆大小，再减/加上s即可
代码
#include <bits/stdc++.h>
using namespace std;
#define umap unordered_map
#define cy cout << "Yes" << endl
#define cn cout << "No" << endl
#define ll long long
#define forn(i, l, r) for (int i = l; i <= r; i++)
#define forn_(i, l, r) for (int i = l; i >= r; i--)
#define debug(a) cout << #a << "=" << a << endl;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 5;
struct cmp1 {
bool operator()(const int &a, const int &b) const {
return a < b;
}
};
struct cmp2 {
bool operator()(const int &a, const int &b) const {
return a > b;
}
};
priority_queue<int, vector<int>, cmp1> q1;
priority_queue<int, vector<int>, cmp2> q2;
ll s1, s2;
ll mid, cnt;
ll q, n;
ll op, x, y;
ll b;
int main() {
cin >> q;
while (q--) {
cin >> op;
if (op == 1) {
cin >> x >> y;
q1.push(x), s1 += x, b += y;
cnt++;
while (q1.size() - q2.size() > 1) {
x = q1.top(), q1.pop();
q2.push(x);
s1 -= x, s2 += x;
}
while (!q2.empty() && q1.top() > q2.top()) {
x = q1.top(), y = q2.top();
q1.pop(), q2.pop();
q2.push(x), q1.push(y);
s1 = s1 - x + y, s2 = s2 - y + x;
}
}
if (op == 2) {
ll res = b;
mid = q1.top();
res += q1.size() * mid - s1;
res += s2 - q2.size() * mid;
cout << mid << ' ' << res << endl;
}
}
return 0;
}
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