Codeforces Round #647 (Div. 2) A - D 解题报告
题目链接
简单写一下题解
\(A. Johnny and Ancient Computer\)
暴力
code
/*
@Author: nonameless
@Date: 2020-06-05 08:12:05
@Email: 2835391726@qq.com
@Blog: https://www.cnblogs.com/nonameless/
*/
#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PLL;
typedef pair<int, int> PII;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f;
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
int main(){
int t; cin >> t;
while(t --){
ll a, b;
cin >> a >> b;
if(a < b) swap(a, b);
int ans = 0;
if(a == b) puts("0");
else if(a % b != 0) puts("-1");
else{
ll t = a / b;
for(int i = 8; i >= 2; i /= 2){
while(t % i == 0){
t /= i;
ans ++;
}
}
if(t > 1) puts("-1");
else cout << ans << endl;
}
}
return 0;
}
\(B. Johnny and His Hobbies\)
集合的数和 \(k\) 去异或得到的集合不变,那么根据异或的特性,我们让集合中的数两两异或,如果有个数出现的次数大于集合里元素的个数,那么就是符合要求的数
code
/*
@Author: nonameless
@Date: 2020-06-05 08:30:50
@Email: 2835391726@qq.com
@Blog: https://www.cnblogs.com/nonameless/
*/
#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PLL;
typedef pair<int, int> PII;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f;
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
int main(){
int t; cin >> t;
while(t --){
int n; cin >> n;
vector<int> a(n + 1);
map<int, int> mp;
for(int i = 1; i <= n; i ++) cin >> a[i];
for(int i = 1; i <= n; i ++)
for(int j = 1; j <= n; j ++){
if(i == j) continue;
int k = a[i] ^ a[j];
mp[k] ++;
}
int ans = -1;
for(auto it : mp){
if(it.second >= n){
ans = it.first;
break;
}
}
cout << ans << endl;
}
return 0;
}
\(C. Johnny and Another Rating Drop\)
打表后我们可以发现规律: \(1, 2, 1, 3, 1, 2, 1, 4...\)。预处理\(2^n\) 处的前缀和,将 \(n\) 转为二进制,求和即可。
code
/*
@Author: nonameless
@Date: 2020-06-05 09:07:15
@Email: 2835391726@qq.com
@Blog: https://www.cnblogs.com/nonameless/
*/
#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PLL;
typedef pair<int, int> PII;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f;
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
const int N = 110;
ll f[N];
ll calc(ll n){
ll res = 0;
string s = "";
while(n){
s += ('0' + n % 2);
n >>= 1;
}
for(int i = 0; i < sz(s); i ++){
if(s[i] == '1') res += f[i];
}
return res;
}
int main(){
f[0] = 1;
for(int i = 1; i < 64; i ++){
f[i] = f[i - 1] * 2 + 1;
}
int t; cin >> t;
while(t --){
ll n; cin >> n;
cout << calc(n) << endl;
}
return 0;
}
\(D. Johnny and Contribution\)
建好图,按将主题小的排序,然后对每一个点进行判断即可。
code
/*
@Author: nonameless
@Date: 2020-06-05 14:02:37
@Email: 2835391726@qq.com
@Blog: https://www.cnblogs.com/nonameless/
*/
#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll;
typedef pair<ll, ll> PLL;
typedef pair<int, int> PII;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f;
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
const int N = 5e5 + 10, M = N << 1;
int idx = 0;
int h[N], to[M], nxt[M];
int col[N];
PII t[N];
int n, m;
void add(int u, int v){
to[ ++ idx] = v; nxt[idx] = h[u]; h[u] = idx;
}
int main(){
cin >> n >> m;
for(int i = 1, u, v; i <= m; i ++){
scanf("%d%d", &u, &v);
add(u, v); add(v, u);
}
for(int i = 1, x; i <= n; i ++){
scanf("%d", &x);
t[i] = {x, i};
col[i] = x;
}
sort(t + 1, t + n + 1);
for(int i = 1; i <= n; i ++){
// cout << t[i].y << " : " << t[i].x << endl;
int u = t[i].y;
set<int> s;
for(int j = h[u]; j; j = nxt[j]){
int v = to[j];
if(col[u] == col[v]) { puts("-1"); return 0; }
s.insert(col[v]);
}
int flag = 1;
for(int j = 1; j < col[u]; j ++){
if(s.find(j) == s.end()){
flag = 0;
break;
}
}
if(!flag) { puts("-1"); return 0; }
}
for(int i = 1; i <= n; i ++)
printf("%d ", t[i].y);
return 0;
}