Educational Codeforces Round 104（A-E题解）
我昨天打了一下Educational Codeforces Round 104的比赛，场内做出了A,B,C,D,E五题，这场的C,D比以往简单。
A. Arena
思路：直接找最小值出现的次数即可。
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int _;
int n;
int a[110];
void solve(){
cin >> _;
while(_--){
cin >> n;
int minn = 1e9;
for(int i = 1; i <= n; i++){
cin >> a[i];
minn = min(minn, a[i]);
}
int ans = 0;
for(int i = 1; i <= n; i++){
if(a[i] == minn){
ans++;
}
}
cout << n - ans << "\n";
}
return ;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
B. Cat Cycle
思路：刚开始被卡了一下，找规律，发现n偶数时候，不会冲突。发现n = 3，会在2，3，4，5，，，，，冲突。n = 5，会在3， 5， 7， ，，，冲突。n = 7，会在4，7，10，，冲突，根据规律推公式即可。
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int _;
ll n, k;
void solve(){
cin >> _;
while(_--){
cin >> n >> k;
if(n % 2 == 0){
cout << 1 + (k - 1) % n << "\n";
}
else{
cout << 1 + (k - 1 + ((k >= (n/2+1)) ? (1 + (k - (n/2+1))/(n/2)) : (0)))%n << "\n";
}
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
C. Minimum Ties
思路：构造题。n是奇数，不需要平局，一个人参加n-1场，一半胜利，一半失败。n是偶数，一个人参加n-1场，n-1此时是奇数，所有要有一场平局。
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int _;
int n;
void solve(){
cin >> _;
while(_--){
cin >> n;
if(n % 2){
int now = n - 1;
int tim = now / 2;
for(int i = 1; i <= n-1; i++){
for(int j = 1; j <= now; j++){
if(j <= tim){
cout << "1" << " ";
}
else{
cout << "-1" << " ";
}
}
now--;
}
}
else{
int now = n - 1;
int tim = now / 2;
for(int i = 1; i <= n-1; i++){
for(int j = 1; j <= now; j++){
if(j <= tim){
cout << "1" << " ";
}
else if(j == tim + 1){
cout << "0" << " ";
}
else{
cout << "-1" << " ";
}
}
now--;
}
}
cout << "\n";
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}
D. Pythagorean Triples
思路：数论题。这个数论比上一场的C简单多了。 1 < = a , b , c < = n 1<=a, b, c <= n 1<=a,b,c<=n范围内找出 a 2 + b 2 = c 2 & & a 2 − b = c a^2+b^2=c^2 \&\&a^2-b=c a2+b2=c2&&a2−b=c的情况个数。
我们先消除 a 2 a^2 a2,得：
c 2 − b 2 = c + b c^2 - b^2 = c+b c2−b2=c+b
c 2 − c = b 2 + b c^2-c=b^2+b c2−c=b2+b
c ∗ ( c − 1 ) = b ( b + 1 ) c*(c-1) = b(b+1) c∗(c−1)=b(b+1)
得出：
c = b + 1 c = b+1 c=b+1
可得出：
a 2 = 2 ∗ b + 1 a^2=2*b+1 a2=2∗b+1
我们找一下 a , b , c a, b, c a,b,c三个数的范围：
1 < = a < = n 1 <= a <= n 1<=a<=n
1 < = b < = n 1 <= b <= n 1<=b<=n
1 < = c < = n 1 <= c <= n 1<=c<=n
1 < = b + 1 < = n 1 <= b+1 <= n 1<=b+1<=n
所以：
1 < = b < = n − 1 1 <= b <= n-1 1<=b<=n−1
此时：
3 < = a 2 < = 2 ∗ n − 1 3 <= a^2 <= 2*n - 1 3<=a2<=2∗n−1
3 < = a < = 2 ∗ n − 1 \sqrt{3} <= a <= \sqrt{2*n-1} 3<=a<=2∗n−1
因为a是整数，我们要找出在 [ 3 , 2 ∗ n − 1 ] [\sqrt{3}, \sqrt{2*n-1}] [3,2∗n−1]中，找出 ( a 2 + 1 ) m o d 2 = = 0 (a^2+1)mod2==0 (a2+1)mod2==0的数的个数，就能得到答案。我们可以通过预处理，前缀和的方式，使得每一个询问复杂度达到 O ( 1 ) O(1) O(1)。
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int _;
const int N = 100010;
ll sum[100010];
ll a[100010];
ll n;
void init(){
memset(a, 0, sizeof a);
memset(sum, 0, sizeof sum);
for(ll i = 2; i <= 100000; i++){
if(((i*i)-1)%2 == 0){
a[i] = 1ll;
}
}
for(int i = 1; i <= 100002; i++){
sum[i] = sum[i-1] + a[i];
}
}
void solve(){
cin >> _;
while(_--){
cin >> n;
ll k = sqrt(2ll*n-1ll);
cout << sum[k] << "\n";
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
init();
solve();
return 0;
}
E. Cheap Dinner
思路：这是一个分层图，我们考虑dp[i][j]表示第i层选第j个的最小花费。第i层对j进行规划的时候，要从上一层和j不冲突的dp中选一个最小值，考虑把dp结合优先队列。
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int _;
ll dp[5][200010];
int n1, n2, n3, n4;
ll a[200010];
ll b[200010];
ll c[200010];
ll d[200010];
set<int> vec[5][200010];
struct cmp{
bool operator () (const pair<ll, int> &a, const pair<ll, int> &b) const{
return a.first > b.first;
}
};
priority_queue<pair<ll, int>, vector< pair<ll, int> >, cmp> q;
void solve(){
cin >> n1 >> n2 >> n3 >> n4;
for(int i = 1; i <= n1; i++){
cin >> a[i];
}
for(int i = 1; i <= n2; i++){
cin >> b[i];
}
for(int i = 1; i <= n3; i++){
cin >> c[i];
}
for(int i = 1; i <= n4; i++){
cin >> d[i];
}
int m1;
cin >> m1;
for(int i = 1; i <= m1; i++){
int x, y;
cin >> x >> y;
vec[2][y].insert(x);
}
int m2;
cin >> m2;
for(int i = 1; i <= m2; i++){
int x, y;
cin >> x >> y;
vec[3][y].insert(x);
}
int m3;
cin >> m3;
for(int i = 1; i <= m3; i++){
int x, y;
cin >> x >> y;
vec[4][y].insert(x);
}
//cout << m1 << m2 << m3 << "\n";
memset(dp, 0x3f, sizeof dp);
for(int i = 1; i <= n1; i++){
dp[1][i] = a[i];
}
for(int tt = 2; tt <= 4; tt++){
//cout << tt << "\n";
if(tt == 2){
while(q.size()){
q.pop();
}
for(int i = 1; i <= n1; i++){
q.push({
dp[1][i], i});
}
//cout << q.size() << "\n";
}
else if(tt == 3){
while(q.size()){
q.pop();
}
for(int i = 1; i <= n2; i++){
q.push({
dp[2][i], i});
}
}
else{
while(q.size()){
q.pop();
}
for(int i = 1; i <= n3; i++){
q.push({
dp[3][i], i});
}
}
vector< pair<ll, int> > tmp;
if(tt == 2){
for(int i = 1; i <= n2; i++){
tmp.clear();
while(q.size() != 0 && vec[tt][i].find((q.top().second)) != vec[tt][i].end()){
pair<ll, int> now = q.top();
q.pop();
tmp.push_back(now);
}
if(q.size() != 0){
pair<ll, int> now = q.top();
dp[tt][i] = min(dp[tt][i], now.first + b[i]);
}
for(int j = 0; j < tmp.size(); j++){
q.push(tmp[j]);
}
}
}
else if(tt == 3){
for(int i = 1; i <= n3; i++){
tmp.clear();
while(q.size() != 0 &&vec[tt][i].find((q.top().second)) != vec[tt][i].end()){
pair<ll, int> now = q.top();
q.pop();
tmp.push_back(now);
}
if(q.size() != 0){
pair<ll, int> now = q.top();
dp[tt][i] = min(dp[tt][i], now.first + c[i]);
}
for(int j = 0; j < tmp.size(); j++){
q.push(tmp[j]);
}
}
}
else{
for(int i = 1; i <= n4; i++){
tmp.clear();
while(q.size() != 0 &&vec[tt][i].find((q.top().second)) != vec[tt][i].end()){
pair<ll, int> now = q.top();
q.pop();
tmp.push_back(now);
}
if(q.size() != 0){
pair<ll, int> now = q.top();
dp[tt][i] = min(dp[tt][i], now.first + d[i]);
}
for(int j = 0; j < tmp.size(); j++){
q.push(tmp[j]);
}
}
}
}
ll ans = 0x3f3f3f3f3f3f3f3f;
for(int i = 1; i <= n4; i++){
ans = min(ans, dp[4][i]);
}
if(ans >= 0x3f3f3f3f3f3f3f3f){
cout << "-1" << "\n";
}
else{
cout << ans << "\n";
}
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
solve();
return 0;
}