比赛地址：freee Programming Contest 2023（AtCoder Beginner Contest 310） - AtCoder
后记：原本写了比较详细的题解，但是，突发意外情况，它没了，所以这份题解略写了。归根结底还是懒
A - Order Something Else
贪心的选价格最小的和优惠券一起使用，最后和原价比大小即可。
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 110;
int n, q, p, mn = 1e9;
int d[N];
int main() {
n = read<int>(), p = read<int>(), q = read<int>();
for (int i = 1, x; i <= n; ++ i) {
x = read<int>();
mn = min(x, mn);
}
cout << min(p, q + mn) << '\n';
return 0;
}
B - Strictly Superior
B - Strictly Superior (atcoder.jp)
按照题意模拟即可。
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 110;
int n, m;
int f[N][N], p[N], c[N];
bitset<N> vis[N];
int main() {
n = read<int>(), m = read<int>();
for (int i = 1; i <= n; ++ i) {
p[i] = read<int>(), c[i] = read<int>();
for (int j = 1; j <= c[i]; ++ j) {
vis[i].set(read<int>());
}
}
for (int i = 1; i <= n; ++ i) {
for (int j = 1; j <= n; ++ j) {
if (i == j) continue ;
if (p[i] == p[j]) {
if (((vis[i] & vis[j]) == vis[i]) && (c[j] > c[i])) {
puts("Yes");
return 0;
}
}
else if (p[i] > p[j]) {
if ((vis[i] & vis[j]) == vis[i]) {
puts("Yes");
return 0;
}
}
}
}
puts("No");
return 0;
}
C - Reversible
字符串哈希来判断，可以搭配 map 来食用。
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 2e5 + 5;
const int based = 29;
int n, len, ans;
ull Hash1[N], Hash2[N];
char s[N];
map<ull, bool> mp;
int main() {
n = read<int>();
for (int i = 1; i <= n; ++ i) {
scanf("%s", s + 1);
len = strlen(s + 1);
Hash1[0] = 0, Hash2[len + 1] = 0;
for (int j = 1; j <= len; ++ j) {
Hash1[j] = Hash1[j - 1] * based + s[j];
}
for (int j = len; j >= 1; -- j) {
Hash2[j] = Hash2[j + 1] * based + s[j];
}
if (mp.count(Hash1[len]) || mp.count(Hash2[1])) {
continue ;
}
mp[Hash1[len]] = 1;
mp[Hash2[1]] = 1;
++ ans;
}
cout << ans << '\n';
return 0;
}
D - Peaceful Teams
D - Peaceful Teams (atcoder.jp)
由于 \(n\) 很小，所以可以直接搜索。
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
int n, t, m, ans;
bool ok[15][15];
vector<int> Team[15];
void dfs(int u) {
if (u == n + 1) {
for (int i = 1; i <= t; ++ i) {
if (!Team[i].size()) {
return ;
}
}
++ ans;
return ;
}
for (int i = 1; i <= t; ++ i) {
int fg = 0;
for (int j : Team[i]) {
fg |= ok[u][j];
}
if (fg) {
continue ;
}
Team[i].emplace_back(u);
dfs(u + 1);
Team[i].pop_back();
if (!Team[i].size()) {
return ;
}
}
}
int main() {
n = read<int>(), t = read<int>(), m = read<int>();
for (int i = 1, x, y; i <= m; ++ i) {
x = read<int>(), y = read<int>();
ok[x][y] = ok[y][x] = 1;
}
dfs(1);
cout << ans << '\n';
return 0;
}
E - NAND repeatedly
E - NAND repeatedly (atcoder.jp)
假设 \(i\) 位置是 \(1\)，则到 \(i - 1\) 位置的值，为 \(1\) 的全变成 \(0\)，为 \(0\) 的全变成 \(1\)，即交换即可，当 \(i\) 位置是 \(0\)，那么所有情况都会变成 \(1\)，最后统计答案即可。
/*
The code was written by yifan, and yifan is neutral!!!
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
template<typename T>
inline T read() {
T x = 0;
bool fg = 0;
char ch = getchar();
while (ch < '0' || ch > '9') {
fg |= (ch == '-');
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return fg ? ~x + 1 : x;
}
const int N = 1e6 + 5;
int n;
ll ans;
string s;
ll a[N], cnt[2];
int main() {
n = read<int>();
for (int i = 1; i <= n; ++ i) {
scanf("%1lld", &a[i]);
}
for (int i = 1; i <= n; ++ i) {
if (a[i]) {
swap(cnt[1], cnt[0]);
}
else {
cnt[1] += cnt[0];
cnt[0] = 0;
}
++ cnt[a[i]];
ans += cnt[1];
}
cout << ans << '\n';
return 0;
}
F 往后没看题，就不写了。
- Contest Programming Beginner AtCoder 报告contest programming beginner atcoder beginner atcoder contest报告 contest programming securities beginner contest programming beginner registry contest programming christmas beginner contest programming beginner keyence contest programming beginner systems beginner atcoder contest 296 beginner atcoder contest 295 beginner atcoder contest abcde