Educational Codeforces Round 38（Div2）

原创 于 2020-11-07 10:36:00 发布 · 109 阅读 · 0 ·
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本文提供了四道算法竞赛题目的解答思路及代码实现，包括字符串处理、最短路径、构造问题和组合数学等方向，旨在帮助读者理解算法原理并掌握实际应用。

A - Word Correction

注意题目中把'y'也算元音 solved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int n;
map<char, int>ma;
char s[N];
stack<char>ss;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    ma['a'] = 1; ma['e'] = 1; ma['i'] = 1; ma['o'] = 1; ma['u'] = 1; ma['y'] = 1;
    n = read();
    scanf("%s", s + 1);

    for (int i = 1; i <= n; ++i) {
        if (ss.empty()) {
            ss.push(s[i]);
        } else {
            char x = ss.top();
            if (ma[x] && ma[s[i]]) continue;
            else {
                ss.push(s[i]);
            }
        }
    }
    stack<char>ans;
    while (!ss.empty()) {
        char x = ss.top();
        ss.pop();
        ans.push(x);
    }

    while (!ans.empty()) {
        char x = ans.top();
        ans.pop();
        cout << x;
    }
    puts("");
    #ifndef ONLINE_JUDGE
        system("pause");
    #endif
}
View Code

B - Run For Your Prize

注意时间不是累加而是取max  solved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll n, ans;
ll a[N];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

ll Min(ll a, ll b, ll c) {
    return min(a, min(b, c));
}

int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    n = read();
    ans = 1e18;
    for (int i = 1; i <= n; ++i) a[i] = read();
    if (n == 1) {
        ans = min(a[1] - 1, 1000000ll - a[1]);
        cout << ans << '\n';
    } else {
        for (int i = 0; i <= n; ++i) {
            ans = min(ans, max(a[i] - 1 , 1000000 - a[i + 1]));
        }
        ans = min(ans, 1000000 - a[1]);
        ans = min(ans, a[n] - 1);
        cout << ans << '\n';
    }
    #ifndef ONLINE_JUDGE
        system("pause");
    #endif
}
View Code

C - Constructing Tests

模几组样例可以发现就是构造满足条件的 n2 - (n / m)2 = x 的n和m 

　可以拆分成(a + b) * (a - b)的形式

赛中找出了规律但是没推出式子

unsolved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 4e5 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll _, x;

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}


int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    _= read();
    while (_--) {
        x = read();
        if (!x) {
            cout << "1 1" << '\n';
            continue;
        }
        // if (x == 1) {
        //     puts("-1");
        //     continue;
        // }
        bool flag = 0;
        for (ll i = 1; i * i <= x; ++i) {
            // cout << i << '\n';
            if (x % i != 0) continue;
            ll xx = x / i, yy = i;
            if ((xx + yy) & 1) continue;

            // printf("%d %d\n", xx, yy);

            ll a = (xx + yy) / 2;
            ll b = (xx - yy) / 2;

            ll n, m;
            n = a;
            if (!b) continue;
            m = n / b;
            // printf("%d %d\n", n, m);
            if (n * n - (n / m) * (n / m) != x) continue;

            cout << n << " " << m << '\n';
            flag = 1;
            break;
            
        }
        if (!flag) puts("-1");
    }
    #ifndef ONLINE_JUDGE
        system("pause");
    #endif
}
View Code

D - Buy a Ticket 

建虚拟源点跑最短路，原图边权值为道路过路费，虚拟点与原图中点的连边为点权，在这种建图模式下松弛操作就和普通迪杰斯特拉相同。

solved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<ll, ll > PII;
priority_queue< PII, vector<PII>, greater<PII> > q;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const ll N = 1e6 + 5, M = 2e6 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll n, m, idx;
ll e[M], ne[M], w[M], h[N];
ll a[N], d[N];
bool vis[N];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

void add(ll a, ll b, ll c) {
    e[idx] = b;
    w[idx] = c;
    ne[idx] = h[a];
    h[a] = idx++;
}

void dij() {
    for (int i = 1; i < N; ++i) d[i] = 1e18;
    q.push({0, 0});
    d[0] = 0;
    while (!q.empty()) {
        PII t = q.top();
        q.pop();
        int u = t.second;
        if (vis[u]) continue;
        vis[u] = 1;
        for (int i = h[u]; ~i; i = ne[i]) {
            ll v = e[i], z = w[i];
            if (d[u] + z < d[v]) {
                d[v] = d[u] + z;
                q.push({d[v], v});
            }
        }
    }
}

int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    n = read(); m = read();
    mm(h, -1);
    for (int i = 1; i <= m; ++i) {
        ll u, v, w;
        u = read(); v = read(); w = read();
        u++; v++;
        w = w * 2;
        add(u, v, w);
        add(v, u, w);
    }
    for (int i = 2; i <= n + 1; ++i) {
        a[i] = read();
        add(0, i, a[i]);
        add(i, 0, a[i]);
    }

    dij();
    for (int i = 2; i <= n + 1; ++i)
        cout << d[i] << " ";
    puts("");

    #ifndef ONLINE_JUDGE
        system("pause");
    #endif
}
View Code

E - Max History

组合数学，太笨了推不出正确的式子

unsolved

#include <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <map> 
#include <stack>
#include <sstream>
#include <set>
// #pragma GCC optimize(2)

//#define int long long
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define rush() int T;scanf("%d",&T);for(int Ti=1;Ti<=T;++Ti)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mm(i,v) memset(i,v,sizeof i);
#define mp(a, b) make_pair(a, b)
#define pi acos(-1)
#define fi first
#define se second
//你冷静一点，确认思路再敲！！！ 

using namespace std;
typedef long long ll;
typedef double db;
typedef pair<int, int > PII;
priority_queue< PII, vector<PII>, greater<PII> > que;
stringstream ssin; //  ssin << string   while ( ssin >> int)
const ll LINF = 0x7fffffffffffffffll;

const int N = 1e6 + 5, M = 4e5 + 5, mod = 1e9 + 7, INF = 0x3f3f3f3f;
ll _, n;
ll a[N];
map<ll, ll>ma;
ll jc[N], ny[N];
bool vis[N];

inline ll read() {
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1; c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0'; c=getchar();}
    return x*f;
}

ll qp(ll a, ll b) {
    ll ans = 1ll;
    while (b) {
        if (b & 1) {
            ans = ans * a % mod;
        }
        a = a * a % mod;
        b = b / 2;
    }
    return ans;
}

void init() { // 预处理阶乘和逆元 
    jc[0] = 1;
    jc[1] = 1;
    ny[1] = 1;
    for (ll i = 2; i <= 1e6; ++i) {
        jc[i] = jc[i - 1] * i % mod;
        ny[i] = qp(jc[i], mod - 2);
    }
}

ll c(ll n, ll m) {
    if (n == 0 || n == m) return 1;
    return jc[m] * ny[n] % mod * ny[m - n] % mod;
}

int main()
{
    // freopen("in.txt","r",stdin);
    // freopen("out.txt","w",stdout);
    init();
    n = read();
    for (int i = 1; i <= n; ++i) a[i] = read();
    sort(a + 1, a + 1 + n);

    for (int i = 2; i <= n; ++i) {
        if (a[i] == a[i - 1]) ma[a[i]] = ma[a[i - 1]];
        else ma[a[i]] = i - 1;
    }

    ll ans = 0;
    for (int i = 1, j = 1; i <= n; i = j + 1) {
        while (a[i] == a[j + 1]) j++;
        if (j == n) break;
        ll m = ma[a[j]];
        // ans += a[i] * m % mod * jc[n] % mod * qp(m, mod - 2) % mod;
        ans += a[i] * (j - i + 1) % mod * jc[n] % mod * qp(n - i + 1, mod - 2) % mod;
        ans %= mod;
    }
    cout << ans << '\n';

    #ifndef ONLINE_JUDGE
        system("pause");
    #endif
}
View Code
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