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&&a2b=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 c2b2=c+b
c 2 − c = b 2 + b c^2-c=b^2+b c2c=b2+b
c ∗ ( c − 1 ) = b ( b + 1 ) c*(c-1) = b(b+1) c(c1)=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=2b+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<=n1
此时:
3 < = a 2 < = 2 ∗ n − 1 3 <= a^2 <= 2*n - 1 3<=a2<=2n1
3 < = a < = 2 ∗ n − 1 \sqrt{3} <= a <= \sqrt{2*n-1} 3 <=a<=2n1
因为a是整数,我们要找出在 [ 3 , 2 ∗ n − 1 ] [\sqrt{3}, \sqrt{2*n-1}] [3 ,2n1 ]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;
}

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