Codeforces Round #586 (Div. 1 + Div. 2)
A. Cards
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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;
}
int n, cnt1, cnt2;
string s;
vector<int> cnt(26);
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)
cnt[x - 'a'] ++ ;
cnt1 = cnt['n' - 'a'];
cnt2 = cnt['z' - 'a'];
for (int i = 1; i <= cnt1; i ++ )
cout << "1 ";
for (int i = 1; i <= cnt2; i ++ )
cout << "0 ";
return 0;
}
B. Multiplication Table
-
思路:(frac{a_xa_y * a_xa_z}{a_ya_z} = {a_x}^2)
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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 = 1e3 + 10;
int n;
ll ans[N];
ll M[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;
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= n; j ++ )
cin >> M[i][j];
ans[1] = sqrt(M[1][2] * M[1][3] / M[2][3]);
for (int i = 2; i <= n; i ++ )
ans[i] = M[1][i] / ans[1];
for (int i = 1; i < n; i ++ )
cout << ans[i] << " ";
cout << ans[n] << "
";
return 0;
}
C. Substring Game in the Lesson
-
思路:瞎猜搞出来的
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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;
}
char min_ = 'z';
string s;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> s;
for (int i = 0; i < s.length(); i ++ ){
if (min_ < s[i])
cout << "Ann
";
else{
min_ = s[i];
cout << "Mike
";
}
}
return 0;
}
D. Alex and Julian
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思路:参照dalao的思路 如果是二分图, 则不存在奇环,.
考虑存在a,则有0->a, a->2a, 如果存在2a,则有0->2a,就存在奇环,如果有4a, 6a 也会存在奇环, 因为0->4a, 0->6a都属于同一边.
考虑存在p, 有0->p, 同时可以存在, 3p, 5p, 考虑p的最高2的幂在p乘上一个奇数以后不会增加, 所以同时只能存在2的幂与p相等的值. 原文链接 -
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 = 70;
int n, m;
ll b[N], x;
vector<ll> a[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 >> n;
for (int i = 1; i <= n; i ++ ){
cin >> b[i];
int tmp = 0;
x = b[i];
while (x % 2 == 0){
x /= 2;
tmp ++ ;
}
a[tmp].push_back(i);
}
m = -1;
for (int i = 0; i <= 64; i ++ )
if (m == -1 || a[m].size() < a[i].size())
m = i;
cout << n - a[m].size() << "
";
for (int i = 0; i <= 64; i ++ )
if (i != m)
for (int j = 0; j < a[i].size(); j ++ )
cout << b[a[i][j]] << " ";
return 0;
}