Codeforces Round #577 (Div. 2)
Codeforces Round #577 (Div. 2)
A. Important Exam
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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;
}
const int N = 1010;
struct Student{
string s;
}stu[N];
int n, m, a, ans;
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;
for (int i = 1; i <= n; i ++ )
cin >> stu[i].s;
for (int i = 0; i < m; i ++ ){
cin >> a;
int cnt[6] = {0};
for (int j = 1; j <= n; j ++ ){
if (stu[j].s[i] == 'A')
cnt[1] ++ ;
else if (stu[j].s[i] == 'B')
cnt[2] ++ ;
else if (stu[j].s[i] == 'C')
cnt[3] ++ ;
else if (stu[j].s[i] == 'D')
cnt[4] ++ ;
else if (stu[j].s[i] == 'E')
cnt[5] ++ ;
}
ans += a * max(cnt[1], max(cnt[2], max(cnt[3], max(cnt[4], cnt[5]))));
}
cout << ans << "\n";
return 0;
}
B. Zero Array
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思路：和为奇数或者和小于最大的数的两倍时 都无法满足
-
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 = 1e5 + 10;
int n;
int a[N];
ll sum;
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;
for (int i = 1; i <= n; i ++ ){
cin >> a[i];
sum += 1ll * a[i];
}
sort(a + 1, a + n + 1);
if (sum & 1 || sum < 2 * a[n])
cout << "NO\n";
else
cout << "YES\n";
return 0;
}
C. Maximum Median
-
思路：二分 以前做过
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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;
}
const int N = 2e5 + 10;
int n, k;
ll ans;
ll a[N];
bool judge(ll x){
ll res = 0;
for (int i = (n + 1) >> 1; i <= n; i ++ )
if (a[i] < x)
res += x - a[i];
if (res <= k)
return true;
return false;
}
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 ++ )
cin >> a[i];
sort(a + 1, a + n + 1);
ll l = a[(n + 1) >> 1], r = 2e9;
while (l <= r){
ll mid = (l + r) >> 1;
if (judge(mid)){
ans = mid;
l = mid + 1;
}
else
r = mid - 1;
}
cout << ans << "\n";
return 0;
}
D. Treasure Hunting
-
思路：二分+dp 状态转移方程还是很好写的 主要是求第j层边界到第i层边界水平移动步数
-
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 = 2e5 + 10;
const int INF = 0x3f3f3f3f;
int n, m, k, q;
ll ans;
ll b[N];
ll a[N][2], dp[N][2];
ll calc(int pos, int u, int i, int v){
ll res = INF, p = lower_bound(b + 1, b + q + 1, a[pos][u]) - b;
if (p <= q)
res = abs(a[pos][u] - b[p]) + abs(a[i][v] - a[i][v ^ 1]) + abs(b[p] - a[i][v ^ 1]);
p = upper_bound(b + 1, b + q + 1, a[pos][u]) - b - 1;
if (p)
res = min(res, abs(a[pos][u] - b[p]) + abs(a[i][v] - a[i][v ^ 1]) + abs(b[p] - a[i][v ^ 1]));
return res;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
for (int i = 1; i < N; i ++ )
a[i][0] = INF, a[i][1] = -INF;
memset(dp, 0x3f, sizeof(dp));
cin >> n >> m >> k >> q;
for (int i = 1; i <= k; i ++ ){
int r, c;
cin >> r >> c;
a[r][0] = min(a[r][0], 1ll * c);
a[r][1] = max(a[r][1], 1ll * c);
};
a[1][0] = 1, a[1][1] = max(a[1][1], 1ll);
dp[1][0] = abs(a[1][1] - 1) + abs(a[1][1] - a[1][0]), dp[1][1] = abs(a[1][1] - 1);
for (int i = 1; i <= q; i ++ )
cin >> b[i];
sort(b + 1, b + q + 1);
int pos = 1;
for (int i = 2; i <= n; i ++ ){
if (a[i][0] == INF)
continue;
for (int v = 0; v < 2; v ++ )
for (int u = 0; u < 2; u ++ )
dp[i][v] = min(dp[i][v], dp[pos][u] + calc(pos, u, i, v) + i - pos);
pos = i;
}
ans = min(dp[pos][0], dp[pos][1]);
cout << ans << "\n";
return 0;
}
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