Codeforces Round #578 (Div. 2)
作者:互联网
Codeforces Round #578 (Div. 2)
A. Hotelier
-
思路:水题
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AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int n;
string s;
int a[10];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> s;
for (auto x: s){
if (x == 'L'){
for (int i = 0; i < 10; i ++ ){
if (!a[i]){
a[i] = 1;
break;
}
}
}
else if (x == 'R'){
for (int i = 9; i >= 0; i -- ){
if (!a[i]){
a[i] = 1;
break;
}
}
}
else
a[x - '0'] = 0;
}
for (int i = 0; i < 9; i ++ )
cout << a[i];
cout << a[9] << "\n";
return 0;
}
B. Block Adventure
-
思路:贪心
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 110;
int t, n, m, k;
int h[N];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
cin >> n >> m >> k;
for (int i = 1; i <= n; i ++ )
cin >> h[i];
for (int i = 1; i < n; i ++ ){
int x = min(h[i], h[i] - (h[i + 1] - k));
m += x;
if (m < 0)
break;
}
if (m < 0)
cout << "NO\n";
else
cout << "YES\n";
}
return 0;
}
C. Round Corridor
-
思路:思维题 圆环被分成了\(gcd(n,m)\)块 判断是不是在同一块即可
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
ll n, m, q, gcd_, sx, sy, ex, ey;
ll calc(ll x, ll y){
if (x == 1)
return y / (n / gcd_);
return y / (m / gcd_);
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
gcd_ = gcd(n, m);
while (q -- ){
cin >> sx >> sy >> ex >> ey;
// cout << calc(sx, sy - 1) << " " << calc(ex, ey - 1) << "\n";
if (calc(sx, sy - 1) == calc(ex, ey - 1))
cout << "YES\n";
else
cout << "NO\n";
}
return 0;
}
D. White Lines
-
思路:比赛时想出来了 没时间写了 赛后十几分钟敲出来了 前缀和处理
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 2010;
int n, k, ans, res;
int x[N][N], y[N][N], cnt_x[N][N], cnt_y[N][N], sum_x[N][N], sum_y[N][N];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> k;
for (int i = 1; i <= n; i ++ ){
string s;
cin >> s;
for (int j = 1; j <= n; j ++ ){
cnt_x[i][j] = cnt_x[i - 1][j] + (s[j - 1] == 'B');
cnt_y[i][j] = cnt_y[i][j - 1] + (s[j - 1] == 'B');
}
}
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n - k + 1; j ++ )
if (cnt_y[i][n] && cnt_y[i][j + k - 1] - cnt_y[i][j - 1] == cnt_y[i][n])
y[i][j] = 1;
for (int i = 1; i <= n - k + 1; i ++ )
for (int j = 1; j <= n; j ++ )
if (cnt_x[n][j] && cnt_x[i + k - 1][j] - cnt_x[i - 1][j] == cnt_x[n][j])
x[i][j] = 1;
for (int i = 1; i <= n; i ++ ){
for (int j = 1; j <= n; j ++ ){
sum_x[i][j] = sum_x[i][j - 1] + x[i][j];
sum_y[i][j] = sum_y[i - 1][j] + y[i][j];
}
}
for (int i = 1; i <= n - k + 1; i ++ )
for (int j = 1; j <= n - k + 1; j ++ )
ans = max(ans, sum_x[i][j + k - 1] - sum_x[i][j - 1] + sum_y[i + k - 1][j] - sum_y[i - 1][j]);
for (int i = 1; i <= n; i ++ )
res += (cnt_x[n][i] == 0) + (cnt_y[i][n] == 0);
// cout << ans << " " << res << "\n";
cout << ans + res << "\n";
return 0;
}
E. Compress Words
-
思路:KMP即可
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int n, len, len_;
string s, t;
void get_nxt(string s, int *nxt){
int j, k;
nxt[0] = -1;
j = 0, k = -1;
while (j < len_){
if (k == -1 || s[j] == s[k]){
j ++ , k ++ ;
nxt[j] = k;
}
else
k = nxt[k];
}
}
int kmp(string s, string t){
int nxt[len_ + 1];
int i, j;
i = j = 0;
get_nxt(t, nxt);
while (i < s.length()){
if (j == -1 || s[i] == t[j])
i ++ , j ++ ;
else
j = nxt[j];
if (j == len_)
return len_;
}
return j;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
cin >> s;
for (int i = 2; i <= n; i ++ ){
cin >> t;
len = s.length(), len_ = t.length();
int len__ = max(0, len - len_);
int tmp = kmp(s.substr(len__), t);
for (int j = tmp; j < len_; j ++ )
s += t[j];
}
cout << s << "\n";
return 0;
}
标签:578,int,res,ll,Codeforces,long,Div,include,mod 来源: https://www.cnblogs.com/Misuchii/p/13917442.html