2020牛客暑期多校訓練第三場部分
阿新 • • 發佈：2020-07-20
L 、 Problem L is the Only Lovely Problem
#include <iostream> #include <cstdio> #include <algorithm> #include <unordered_map> #include <vector> #include <map> #include <list> #include <queue> #include <cstring> #include <cstdlib> #include <ctime> #include <cmath> #include <stack> #include <set> #pragma GCC optimize(3 , "Ofast" , "inline") using namespace std ; #define ios ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) #define x first #define y second typedef long long ll ; const double esp = 1e-6 , pi = acos(-1) ; typedef pair<int , int> PII ; const int N = 1e6 + 10 , INF = 0x3f3f3f3f , mod = 1e9 + 7; ll in() { ll x = 0 , f = 1 ; char ch = getchar() ; while(!isdigit(ch)) {if(ch == '-') f = -1 ; ch = getchar() ;} while(isdigit(ch)) x = x * 10 + ch - 48 , ch = getchar() ; return x * f ; } int main() { string s ; cin >> s ; string a = "lovely" ; for(int i = 0 ;i < 6 ;i ++ ) { if(i >= s.size()) break ; if(s[i] < 'a') s[i] += 32 ; if(s[i] != a[i]) return 0 * puts("ugly") ; } return 0 * puts("lovely") ; } /* */
B、Classical String Problem
#include <iostream> #include <cstdio> #include <algorithm> #include <unordered_map> #include <vector> #include <map> #include <list> #include <queue> #include <cstring> #include <cstdlib> #include <ctime> #include <cmath> #include <stack> #include <set> #pragma GCC optimize(3 , "Ofast" , "inline") using namespace std ; #define ios ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) #define x first #define y second typedef long long ll ; const double esp = 1e-6 , pi = acos(-1) ; typedef pair<int , int> PII ; const int N = 2e6 + 10 , INF = 0x3f3f3f3f , mod = 1e9 + 7; ll in() { ll x = 0 , f = 1 ; char ch = getchar() ; while(!isdigit(ch)) {if(ch == '-') f = -1 ; ch = getchar() ;} while(isdigit(ch)) x = x * 10 + ch - 48 , ch = getchar() ; return x * f ; } char s[N] ; int main() { scanf("%s" , s) ; int n = in() ; int len = strlen(s) ; int l = 0 , r = len - 1 ; for(int i = 1; i <= n ;i ++ ) { char c[2] ; int x ; scanf("%s%d" , c , &x) ; if(c[0] == 'A') printf("%c\n" , s[((l + x - 1) % len + len) % len]) ; else l += x , r += x , (l %= len + len) %= len , (r %= len + len) %= len ; } return 0 ; } /* */
A、Clam and Fish
#include <iostream> #include <cstdio> #include <algorithm> #include <unordered_map> #include <vector> #include <map> #include <list> #include <queue> #include <cstring> #include <cstdlib> #include <ctime> #include <cmath> #include <stack> #include <set> #pragma GCC optimize(3 , "Ofast" , "inline") using namespace std ; #define ios ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0) #define x first #define y second typedef long long ll ; const double esp = 1e-6 , pi = acos(-1) ; typedef pair<int , int> PII ; const int N = 2e6 + 10 , INF = 0x3f3f3f3f , mod = 1e9 + 7; char s[N] ; void work(){ int n ; scanf("%d" , &n) ; int ans = 0 , res = 0 ; scanf("%s" , s) ; for(int i = 0 ;i < n ;i ++) { if(s[i] >= '2') { ans ++ ; } if(s[i] == '0') { if(res) res -- , ans ++ ; } if(s[i] == '1') res ++ ; } printf("%d\n" , ans + res / 2) ; return ; } int main() { int t ; scanf("%d" , &t) ; while(t --) work() ; return 0 ; } /* */
c、Operation Love
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <unordered_map>
#include <vector>
#include <map>
#include <list>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <stack>
#include <set>
#pragma GCC optimize(3 , "Ofast" , "inline")
using namespace std ;
#define ios ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0)
#define x first
#define y second
typedef long long ll ;
const double esp = 1e-5 , pi = acos(-1) ;
typedef pair<double , double> PII ;
const int N = 1e6 + 10 , INF = 0x3f3f3f3f , mod = 1e9 + 7;
ll in()
{
ll x = 0 , f = 1 ;
char ch = getchar() ;
while(!isdigit(ch)) {if(ch == '-') f = -1 ; ch = getchar() ;}
while(isdigit(ch)) x = x * 10 + ch - 48 , ch = getchar() ;
return x * f ;
}
PII a[N] ;
int len(PII a , PII b){
double x = a.x - b.x , y = a.y - b.y ;
return sqrt(x * x + y * y + 0.5) ;
}
double cross(PII a , PII b){
return a.x * b.y - a.y * b.x ;
}
double area() {
double ans = 0 ;
for(int i = 0 ;i < 20 ;i ++)
ans += cross(a[i] , a[(i + 1) % 20]) ;
return ans ;
}
void work()
{
vector<int> ans ;
for(int i = 0; i < 20 ;i ++ ) cin >> a[i].x >> a[i].y ;
if(area() > 0) {
for(int i = 0; i < 20 ;i ++) {
if(len(a[i] , a[(i + 1) % 20]) == 9) {
if(len(a[(i + 1) % 20] , a[(i + 2) % 20]) == 8) cout << "right" << endl ;
else cout << "left" << endl ;
return ;
}
}
}else {
for(int i = 0; i < 20 ;i ++) {
if(len(a[i] , a[(i + 1) % 20]) == 9) {
if(len(a[(i + 1) % 20] , a[(i + 2) % 20]) == 8) cout << "left" << endl ;
else cout << "right" << endl ;
return ;
}
}
}
return ;
}
int main()
{
int n = in() ;
while(n --) work() ;
return 0 ;
}
/*
*/
E、Two Matchings
首先
\[p_{p[i]} = i, 那麼p_i = p[i]\\例如p_5 = 3,那麼就是i = 3, p_3 = 5 \]
根據上述，發現一個目標n的排列裡面，都是一個一個的環，並且每個環的長度為2 ， 題目要求兩個那個matching串和最小，就貪心使每個串的和比較小，也就是找個最小和次小，那麼第一個matching最小的話，肯定是直接將a排序，a[1] < a[2] < a[3] < a[4] , 這個貢獻就是a[i + 1] - a[i] , i += 2， 那麼次小的怎麼算呢
\[a[1] < a[2] < a[3] < a[4] < a[5] < a[6] < a[7] < a[8] \]
\[最小的安排是a[2] - a[1] , a[4] - a[3] , a[6] - a[5] \]
貪心講：次小的應該是錯開一個減，例如a[3] - a[1] , a[4] - a[2],但是列舉6個時候應該是a[6] - a[4] ,. a[5] - a[2] , a[3] - a[1]， 如下圖， 那麼8個時候怎麼錯開，發現就是兩個4錯開，10個時候錯開的話就是一個4和一個6組合，下面也就是dp了，每次dp就max一下上面兩種錯開方式的最大值
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <unordered_map>
#include <vector>
#include <map>
#include <list>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <stack>
#include <set>
#pragma GCC optimize(3 , "Ofast" , "inline")
using namespace std ;
#define ios ios::sync_with_stdio(false) , cin.tie(0) , cout.tie(0)
#define x first
#define y second
typedef long long ll ;
const double esp = 1e-6 , pi = acos(-1) ;
typedef pair<int , int> PII ;
const int N = 1e6 + 10 , INF = 0x3f3f3f3f , mod = 1e9 + 7;
ll in()
{
ll x = 0 , f = 1 ;
char ch = getchar() ;
while(!isdigit(ch)) {if(ch == '-') f = -1 ; ch = getchar() ;}
while(isdigit(ch)) x = x * 10 + ch - 48 , ch = getchar() ;
return x * f ;
}
ll a[N] , dp[N] ;
void work(){
int n = in() ;
for(int i = 1; i <= n ;i ++ ) a[i] = in() , dp[i] = 1e18 ;
sort(a + 1 , a + n + 1) ;
ll ans = 0 ;
for(int i = 1; i <= n ;i += 2) {
ans += a[i + 1] - a[i] ;
}
dp[0] = 0 ;
for(int i = 4; i <= n ;i ++ ) {
dp[i] = min(dp[i] , dp[i - 4] + a[i] - a[i - 2] + a[i - 1] - a[i - 3]) ;
if(i >= 6)
dp[i] = min(dp[i] , dp[i - 6] + a[i] - a[i - 2] + a[i - 1] - a[i - 4] + a[i - 3] - a[i - 5]) ;
}
printf("%lld\n" , dp[n] + ans) ;
return ;
}
int main()
{
int n = in() ;
while(n --) work() ;
return 0 ;
}
/*
*/