Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2)
__Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) __
A. Forgetting Things
-
思路：\(a\)和\(b\)相等或者\(a + 1 = b\)或者\(a = 9\) \(b = 1\)才可
-
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 a, b, x;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> a >> b;
if (a == 9 && b == 1){
cout << "9 10\n";
return 0;
}
if (a == b)
cout << a << "0 " << b << "1\n";
else if (a + 1 == b)
cout << a << " " << b << "\n";
else
cout << -1 << "\n";
return 0;
}
B. TV Subscriptions
-
思路：滑动窗口
辣鸡评测鸡 昨晚上交system test给我t了22 结果我早上交相同的代码就过了 -
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 M = 1e6 + 10;
const int INF = 0x3f3f3f3f;
int t, n, k, d, min_, ans;
int a[N], b[M];
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 -- ){
memset(b, 0, sizeof(b));
cin >> n >> k >> d;
ans = 0;
for (int i = 1; i <= n; i ++ )
cin >> a[i];
for (int i = 1; i <= d; i ++ )
if (b[a[i]] ++ == 0)
ans ++ ;
min_ = ans;
for (int i = d + 1; i <= n; i ++ ){
if (( -- b[a[i - d]]) == 0)
ans -- ;
if ((b[a[i]] ++ ) == 0)
ans ++ ;
if (ans < min_)
min_ = ans;
}
cout << min_ << "\n";
}
return 0;
}
C. p-binary
- 思路：从1开始枚举ans 统计\(n - ans * p\)二进制中一的个数即可
昨晚上没想到__builtin_popcountll_ 睡醒早上才想到 - 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, p, ans = 1;
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 >> p;
while (n - ans * p > 0){
if (__builtin_popcountll(n - ans * p) <= ans){
if (n - ans * p >= ans){
cout << ans << "\n";
return 0;
}
}
ans ++ ;
}
cout << "-1\n";
return 0;
}
D. Power Products
-
思路：照着官方题解写的 先将\(a\)唯一分解定理 对于每一项\(bi ^ ki\) 当且仅当\(ki % k == 0\)时才对结果有贡献 最后统计下即可
-
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(ll a, ll b){
ll res = 1;
while (b){
if (b & 1)
res *= a;
a *= a;
b >>= 1;
}
return res;
}
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, k, a;
ll ans;
map<vector<pair<int, int> >, int > mp;
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 ++ ){
vector<pair<int, int> > vec, res;
cin >> a;
for (int b = 2; b * b <= a; b ++ ){
int k_ = 0;
while (a % b == 0){
k_ ++ ;
a /= b;
}
if (k_ % k)
vec.push_back(make_pair(b, k_ % k));
}
if (a > 1)
vec.push_back(make_pair(a, 1));
for (int j = 0; j < vec.size(); j ++ )
res.push_back(make_pair(vec[j].first, k - vec[j].second));
ans += mp[res];
mp[vec] ++ ;
}
cout << ans << "\n";
return 0;
}
