Educational Codeforces Round 24
Educational Codeforces Round 24
https://codeforces.com/contest/818
有些题就是从某个角度想好复杂，不好实现，但是换一种思考方式，从另一个角度想就会豁然开朗，也很好写。这么讲很抽象，但是做题的时候会发现，自己是在一方向上耗死了，结果一看题解就觉得，哇塞tql
A. Diplomas and Certificates
#include <bits/stdc++.h>
#define ll long long
using namespace std;
int main () {
ll n, m, k;
cin >> n >> k;
m = n;
n /= 2;
ll x = n / (k + 1);
if ((k + 1) * x > n) x = 0;
cout << x << ' ' << k * x << ' ' << m - (k + 1) * x;
}
B. Permutation Game
差分构造
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 105;
int a[N], b[N], n, m;
int main () {
cin >> n >> m;
for (int i = 1; i <= m; i++) cin >> b[i];
set<int> s; //未放的数字
for (int i = 1; i <= n; i++) s.insert (i);
for (int i = 1; i < m; i++) {
int idx = b[i], dx = (b[i+1] - b[i] + n) % n;
if (dx == 0) dx = n;
if ((a[idx] && a[idx] != dx) || (!a[idx] && !s.count (dx))) {
cout << -1;
return 0;
}
s.erase (dx), a[idx] = dx;
//cout << idx << ' ' << dx << endl;
}
for (int i = 1; i <= n; i++) {
if (a[i] == 0) a[i] = *s.begin (), s.erase (s.begin ());
cout << a[i] << ' ';
}
}
C. Sofa Thief
D. Multicolored Cars
这个就是要换一个角度去想，记录不可能来check
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5 + 5;
int a[N], n, A;
int main () {
scanf ("%d%d", &n, &A);
for (int i = 1; i <= n; i++) scanf ("%d", &a[i]);
map<int, int> mp;
set<int> s;
for (int i = 1; i <= 1e6; i++) s.insert (i);
s.erase (A);
for (int i = 1; i <= n; i++) {
mp[a[i]] ++;
if (a[i] != A) {
if (mp[a[i]] <= mp[A]) s.erase (a[i]);
}
}
for (auto i : s) {
if (mp[i] >= mp[A]) {
printf ("%d", i);
return 0;
}
}
printf ("-1");
}
//每个阶段可能的答案取个交集
E. Card Game Again
半暴力优化查找
再已知确实有答案(前/后缀乘积)的情况下,找到最小区间
记录上一个位置防止重复查找
特别关注循环边界的退出条件！！
#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 1e5 + 5;
ll a[N], n, k;
int main () {
cin >> n >> k;
ll ans = 0, lst = n, mul = 1;
for (int i = 1; i <= n; i++) cin >> a[i];
for (int i = n, j; i >= 1; i--) {
mul = mul * a[i] % k;
if (mul) continue;
mul = 1;
for (j = i; mul * a[j] % k; ) {
mul = mul * a[j++] % k;
if (mul * a[j] % k == 0) { //最小区间[i,j]
break;
}
}
ans += i * (lst - j + 1);
lst = j - 1;
}
cout << ans << endl;
}
//半暴力优化查找
//再已知确实有答案(前/后缀乘积)的情况下,找到最小区间
//记录上一个位置防止重复查找
