Educational Codeforces Round 129 (Rated for Div. 2) A - D 题解

传送门

A. Game with Cards

看最大的在谁那，谁就赢

如果最大的都一样，则先手赢

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define endl '\n'
#define pii pair
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while(t--)
    {
        int a = 0;
        int n;
        cin >> n;
        for(int i=0; i> x; a = max(a, x);}
        int m;
        cin >> m;
        int b = 0;
        for(int i=0; i> x; b = max(b, x);}
        if(b > a) cout << "Bob" << endl;
        else cout << "Alice" << endl;
        if(a > b) cout << "Alice" << endl;
        else cout << "Bob" << endl;
    }
    return 0;
}

B. Card Trick

它的移动并不会改变牌之间的相对位置

也就是相当于把牌看做一个环，有一个指针指向第一个，如果洗了 n 张牌，则指针向后了 n 位

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define endl '\n'
#define pii pair
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;
int num[maxn];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while(t--)
    {
        int n;
        cin >> n;
        for(int i=0; i> num[i];
        ll tp = 0;
        int m;
        cin >> m;
        for(int i=0; i> x;
            tp += x;
        }
        tp %= n;
        cout << num[tp] << endl;
    }
    return 0;
}

C. Double Sort

排序后的位置是唯一的，所以考虑排序后，a b 数组原来的位置能不能同时换到排序后的位置 if(a[j] == aa[i] && b[j] == bb[i] && vis[j] == 0，如果所有数字都能，则证明有解

然后就选择排序一样，把原来的 a b 数组排到排好序的位置上，输出答案即可

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
#define endl '\n'
#define pii pair
const ll maxn = 110;
const ll inf = 1e17 + 10;
int a[maxn], b[maxn], aa[maxn], bb[maxn];
int vis[maxn];
int num[maxn];

int main()
{
    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
    int t;
    cin >> t;
    while(t--)
    {
        int n;
        cin >> n;
        for(int i=1; i<=n; i++) {cin >> a[i]; aa[i] = a[i];}
        for(int i=1; i<=n; i++) {cin >> b[i]; bb[i] = b[i];}
        for(int i=1; i<=n; i++) vis[i] = 0;
        sort(aa + 1, aa + n + 1);
        sort(bb + 1, bb + n + 1);
        int f = 1;
        for(int i=1; i<=n && f; i++)
        {
            int way = 0;
            for(int j=1; j<=n; j++)
            {
                if(a[j] == aa[i] && b[j] == bb[i] && vis[j] == 0)
                {
                    way = j;
                    break;
                }
            }
            if(way == 0) f = 0;
            else vis[way] = i;
        }
        if(f == 0) cout << -1 << endl;
        else
        {
            vectorans;
            for(int i=1; i<=n; i++)
            {
                int way = 0;
                for(int j=i; j<=n; j++)
                {
                    if(vis[j] == i)
                    {
                        way = j;
                        break;
                    }
                }
                if(i != way)
                {
                    ans.push_back(make_pair(i, way));
                    vis[way] = vis[i];
                }
            }
            cout << ans.size() << endl;
            for(int i=0; i


D. Required Length
bfs 直接搜就好了
一开始还以为会爆 unsigned long long，在接近边界的时候还特判了一下
结果发现是小学数学没学好
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef unsigned long long ll;
#define endl '\n'
#define pii pair
const ll maxn = 110;
const ll inf = 1e17 + 10;
mapvis;
ll n, m, nn = 1;
ll cnt[11];

int bfs(ll s)
{
    queueq;
    q.push(make_pair(s, 0));
    while(q.size())
    {
        pii now = q.front();
        q.pop();
        if(now.first == 0) return now.second;
        if(vis[now.first]) continue;
        vis[now.first] = now.second;
        ll t = now.first, len = 0;
        while(t)
        {
            cnt[t % 10] = 1;
            t /= 10;
            len++;
        }
        if(len >= n) return now.second;
        for(int i=2; i<=9; i++)
            if(cnt[i])
                q.push(make_pair(now.first * i, now.second + 1));
        for(int i=0; i<=9; i++) cnt[i] = 0;
    }
    return -1;
}

int main()
{
    // ios::sync_with_stdio(false);
    // cin.tie(0);
    // cout.tie(0);
    cin >> n >> m;
    for(int i=0; i


					
					
					
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