AtCoder Beginner Contest 310
A - Order Something Else
#include <bits/stdc++.h>
using namespace std;
#define int long long
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int n , p , q;
cin >> n >> p >> q;
int res = p;
for( int x ; n ; n -- ){
cin >> x , res = min( res , q + x );
}
cout << res << "\n";
return 0;
}
B - Strictly Superior
#include <bits/stdc++.h>
using namespace std;
#define int long long
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int n, m;
cin >> n >> m;
vector<int> p(n + 1);
vector<vector<int>> f(n + 1, vector<int>(m + 1));
for (int i = 1, c; i <= n; i++) {
cin >> p[i] >> c;
for (int x; c; c--)
cin >> x, f[i][x] = 1;
}
for( int i = 1 ; i <= n ; i ++ ){
for( int j = 1 ; j <= n ; j ++ ){
if( p[i] < p[j] ) continue;
int q = 1 , t = 0 ;
for( int l = 1 ; q && l <= m ; l ++ ){
if( f[j][l] > f[i][l] ) t ++;
else if( f[j][l] < f[i][l] ) q = 0;
}
if( q == 0 ) continue;
if( p[i] > p[j] || t > 0 ){
cout << "Yes\n";
return 0;
}
}
}
cout << "No\n";
return 0;
}
C - Reversible
只存字典序较小的
#include <bits/stdc++.h>
using namespace std;
#define int long long
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int n;
cin >> n;
set<string> cnt;
for (string s, t; n; n--) {
cin >> s, t = s;
reverse(s.begin(),s.end());
s = min( s , t );
cnt.insert(s);
}
cout << cnt.size() << "\n";
return 0;
}
D - Peaceful Teams
直接暴搜，注意在搜索的过程中要及时盼到不合法的情况
#include <bits/stdc++.h>
using namespace std;
#define int long long
int n, T, m, res;
vector<vector<int>> g;
vector<set<int>> s;
void dfs(int x) {
if (x > n) {
if( s.size() == T ) res ++;
return;
}
if( s.size() + n - x + 1 < T ) return;
if (s.size() < T) {
s.push_back(set<int>());
s.back().insert(x);
dfs(x + 1);
s.pop_back();
}
for (auto &i: s) {
int f = 1;
for (auto j: g[x])
if (i.count(j)) {
f = 0;
break;
}
if( f == 0 ) continue;
i.insert(x);
dfs( x + 1 );
i.erase(x);
}
return;
}
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
cin >> n >> T >> m;
g = vector<vector<int>>(n + 1);
for (int x, y; m; m--) {
cin >> x >> y;
if (x < y) swap(x, y);
g[x].push_back(y);
}
dfs(1);
cout << res << "\n";
return 0;
}
E - NAND repeatedly
简单的做一个 dp 就好了。
#include <bits/stdc++.h>
using namespace std;
#define int long long
int32_t main() {
ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
int n , res = 0 ;
string s;
cin >> n >> s;
vector<array<int,2>> f(n+1);
for( int i = 1 , x ; i <= n ; i ++ ){
x = s[i-1] - '0';
if( x == 1 ) {
f[i][1] = 1 + f[i-1][0];
f[i][0] = f[i-1][1];
}else{
f[i][1] = f[i-1][0] + f[i-1][1];
f[i][0] = 1;
}
res += f[i][1];
}
cout << res << '\n';
return 0;
}
F - Make 10 Again
状压 dp，\(f[i][s]\)表示前\(i\)个数选部分数可以构成集合的概率。
对于枚举除了的\(s\)以及当前骰子掷出的数字\(j\)，有三种情况
s表示当前不要1<<(j-1)表示之前的都不要s<<j表示之前的加上当前的
把上述三种情况与到一起就是当前的数字全部可以构成的集合，注意要与\((1<<10)-1\)与一下防止溢出。
最后枚举所有包含10的集合把概率求和即为答案
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 998244353;
const int N = 1 << 10;
int power(int x, int y) {
int ans = 1;
while (y) {
if (y & 1) ans = ans * x % mod;
x = x * x % mod, y >>= 1;
}
return ans;
}
int inv(int x) {
return power(x, mod - 2);
}
int read() {
int x = 0, f = 1, ch = getchar();
while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
if (ch == '-') f = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
return x * f;
}
int32_t main() {
int n = read();
vector<int> f(N);
f[0] = 1;
for (int x, y, invs; n; n--) {
x = read(), y = min(10ll, x), invs = inv(x);
vector<int> g(N);
for (int s = 0; s < N; s++) {
for (int i = 1; i <= y; i++)
(g[(s | s << i | 1 << (i - 1)) & (N - 1)] += f[s] * invs % mod) %= mod;
(g[s] += f[s] * (x - y) % mod * invs % mod) %= mod;
}
swap(f, g);
}
int res = 0;
for (int i = (1 << 9); i < N; i++)
(res += f[i]) %= mod;
cout << res << "\n";
return 0;
}
