Codeforces Round #579 (Div. 3)
A. Circle of Students
这题我是直接把正序、逆序的两种放在数组里面直接判断。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 205;
int q, n;
int a[N], b[N], c[N];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> q;
while(q--) {
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
int p;
for(int i = 1; i <= n; i++) {
if(a[i] == 1) {
p = i; break;
}
}
for(int i = p; i <= n; i++) {
b[i - p + 1] = a[i];
}
for(int i = 1; i < p; i++) {
b[i + n - p + 1] = a[i];
}
for(int i = p; i >= 1; i--) {
c[p - i + 1] = a[i];
}
for(int i = n; i > p; i--) {
c[n - i + p + 1] = a[i];
}
int ok = 1;
for(int i = 1; i <= n; i++) {
if(b[i] - b[i - 1] != 1) ok = 0;
}
if(ok) {
cout << "YES" << '
';
continue;
}
ok = 1;
for(int i = 1; i <= n; i++) {
if(c[i] - c[i - 1] != 1) ok = 0;
}
if(ok) {
cout << "YES" << '
';
continue;
}
cout << "NO" << '
';
}
return 0;
}
B. Equal Rectangles
排序后判断即可,满足两个条件:对应边长相等;面积乘积相等。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1005;
int q, n;
int a[N], b[N];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> q;
while(q--) {
cin >> n; n <<= 2;
for(int i = 1; i <= n; i++) cin >> a[i];
sort(a + 1, a + n + 1);
int i = 1, j = n;
int ok = 1;
int ans = -1;
while(i < j) {
if(a[i] == a[i + 1] && a[j] == a[j - 1]) {
if(ans == -1) {
ans = a[i] * a[j];
} else {
if(ans != a[i] * a[j]) {
ok = 0; break;
}
}
} else {
ok = 0; break;
}
i += 2; j -= 2;
}
if(ok) cout << "YES" << '
';
else cout << "NO" << '
';
}
return 0;
}
C. Common Divisors
水题。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 4e5 + 5;
int q, n;
ll a[N];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
ll g = a[1];
for(int i = 2; i <= n; i++) {
g = __gcd(g, a[i]);
}
int ans = 0;
for(int i = 1; 1ll * i * i <= g; i++) {
if(g % i == 0) {
ans ++;
if(g / i != i) ans++;
}
}
cout << ans;
return 0;
}
D2. Remove the Substring (hard version)
对于串(t)中的每个位置,求出其在串(s)中最早能出现的位置以及最晚能出现的位置。之后逐一枚举判断即可。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 5;
char s[N], t[N];
int f[N][26], g[N][26];
int nxt[26];
int h[N][2];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> s + 1 >> t + 1;
int n = strlen(s + 1), m = strlen(t + 1);
for(int i = n; i >= 0; i--) {
for(int j = 0; j < 26; j++) f[i][j] = nxt[j];
nxt[s[i] - 'a'] = i;
}
for(int i = 0; i < 26; i++) nxt[i] = 0;
for(int i = 1; i <= n + 1; i++) {
for(int j = 0; j < 26; j++) g[i][j] = nxt[j];
nxt[s[i] - 'a'] = i;
}
int now = 0;
for(int i = 1; i <= m; i++) {
h[i][0] = f[now][t[i] - 'a'];
now = f[now][t[i] - 'a'];
}
now = n + 1;
for(int i = m; i >= 1; i--) {
h[i][1] = g[now][t[i] - 'a'];
now = g[now][t[i] - 'a'];
}
int res = max(h[1][1] - 1, n - h[m][0]);
// cout << h[1][1] << '
';
for(int i = 1; i < m; i++) {
res = max(res, h[i + 1][1] - h[i][0] - 1);
}
cout << res;
return 0;
}
E. Boxers
用个桶来进行标记就行,对于一个数,优先标记其前面的那一个数,其次标记本身,最后标记后面。
因为我们是从小到大枚举,可以证明这样最优。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 150005;
int q, n;
int a[N], c[N];
bool vis[N];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> n;
for(int i = 1; i <= n; i++) cin >> a[i];
for(int i = 1; i <= n; i++) c[a[i]]++;
if(c[1] > 1) {
vis[1] = vis[2] = 1;
} else if(c[1] == 1) vis[1] = 1;
for(int i = 2; i < N; i++) {
if(c[i]) {
if(!vis[i - 1]) vis[i - 1] = 1;
else if(!vis[i]) vis[i] = 1;
else vis[i + 1] = 1;
if(c[i] == 2) {
if(!vis[i - 1] && i > 1) vis[i - 1] = 1;
else if(!vis[i]) vis[i] = 1;
else vis[i + 1] = 1;
}
if(c[i] >= 3) {
vis[i] = 1;
vis[i - 1] = 1;
vis[i + 1] = 1;
}
}
}
int ans = 0;
for(int i = 1; i < N; i++) ans += vis[i];
cout << ans;
return 0;
}
F1. Complete the Projects (easy version)
首先考虑把所有(y>0)的给搞完,然后对于(y<0)的,按照(x+y)从大到小排序,再逐一判断就行。
这里从大到小排序可以保证每次能够选得尽量地多,就相当于处理一个(r->a_i+r)的逆问题。
Code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 60005;
int q, n, m, r;
struct node{
int r, v;
}a[N], b[N];
bool vis[N];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> n >> r;
for(int i = 1; i <= n; i++) {
cin >> a[i].r >> a[i].v;
}
while(1) {
int mx = -N, id;
for(int i = 1; i <= n; i++) {
if(!vis[i] && a[i].r <= r && a[i].v > mx) {
mx = a[i].v; id = i;
}
}
if(mx < 0) break;
vis[id] = 1;
r += mx;
}
int tot = 0;
for(int i = 1; i <= n; i++) {
if(!vis[i]) b[++tot] = a[i];
}
sort(b + 1, b + tot + 1, [&](node a, node b){return a.r + a.v > b.r + b.v;});
int ok = 1;
for(int i = 1; i <= tot; i++) {
if(r >= b[i].r && r + b[i].v >= 0) {
r += b[i].v;
} else ok = 0;
}
if(ok) cout << "YES" << '
';
else cout << "NO" << '
';
return 0;
}
F2. Complete the Projects (hard version)
类似于之前的思路,首先贪心地把(y>0)的部分尽量取完,然后对于(y<0)的(dp)来处理。
设(dp[i][j])表示当前(i)个物品中选了若干个,(r)为(j)时取的最大数量,转移时类似于背包。之后可以滚动数组优化一维。
注意一下我们还是要按照(x+y)来排序,否则有些状态不能转移到。
Code
#include <bits/stdc++.h>
#define MP make_pair
#define fi first
#define se second
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 105, MAX = 60005;
int n, r;
pii a[N], b[N];
bool vis[N];
int dp[MAX];
int main() {
ios::sync_with_stdio(false); cin.tie(0);
cin >> n >> r;
for(int i = 1; i <= n; i++) cin >> a[i].fi >> a[i].se;
sort(a + 1, a + n + 1);
int ans = 0;
while(1) {
int mx = -MAX, id = -1;
for(int i = 1; i <= n; i++) {
if(!vis[i] && a[i].fi <= r && a[i].se > mx) {
mx = a[i].se; id = i;
}
}
if(mx < 0) break;
vis[id] = 1;
r += mx; ans++;
}
int tot = 0;
for(int i = 1; i <= n; i++) {
if(a[i].fi <= r && a[i].se < 0) b[++tot] = a[i];
}
sort(b + 1, b + tot + 1, [&](pii a, pii b){
return a.fi + a.se == b.fi + b.se ? a.fi > a.se : a.fi + a.se > b.fi + b.se;
});
// for(int i = 1; i <= tot; i++) cout << b[i].fi << ' ' << b[i].se << '
';
for(int i = 1; i <= tot; i++) {
for(int j = b[i].fi; j <= r; j++) {
if(j + b[i].se >= 0) dp[j + b[i].se] = max(dp[j + b[i].se], dp[j] + 1);
}
}
int res = 0;
for(int i = 0; i <= r; i++) res = max(res, dp[i]);
ans += res;
cout << ans;
return 0;
}