Educational Codeforces Round 79 (Rated for Div. 2)

题库链接

https://codeforces.com/contest/1279

A. New Year Garland

红色,绿色,蓝色的灯,相同颜色的不能相邻,问能不能连起来
选出颜色最多的那个灯a,然后最少必须有a-1个其他的灯,把灯间隔开来

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 200100;
const int M = 1e9+7;
int n,m,k,t;
int a[4];
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin(t);
    while(t--)
    {
        cin>>a[0]>>a[1]>>a[2];
        sort(a,a+3);
        if(a[0]+a[1] >= a[2]-1)
        {
            puts("YES");
        }
        else puts("NO");
    }
    return 0;
}

B. Verse For Santa

按顺序背诵诗歌,一共有s秒的时间,可以跳过一次背诵,问跳过哪次背诵可以使总的背诵更多
模拟一下就好了,当时间大于s的时候,减去前面最长的背诵,如果小于s,就可以使得总的背诵更多

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 100100;
const int M = 1e9+7;
int n,s,t;
 
int a[maxn];
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin(t);
    while(t--)
    {
        cin(n);cin(s);
        for(int i = 0; i < n; i++) 
        {
            cin(a[i]);
        }
        ll sum = 0;
        ll idx = 0;
        ll mx = 0;
        for(int i = 0; i < n; i++) 
        {
            sum += a[i];
            if(a[i] > mx)
            {
                mx = a[i];
                idx = i;
            }
            if(sum > s)
            {
                if(sum-mx > s) idx = -1;
                else break;
            }
        }
        if(sum <= s) idx = -1;
        printf("%d\n",idx+1);
    }
    return 0;
}

C. Stack of Presents

模拟

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 100100;
const int M = 1e9+7;
int n,m,t;
 
int a[maxn];
int b[maxn];
bool vis[maxn];
 
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    cin>>t;
    while(t--)
    {   
        mem(vis,0);
        cin>>n>>m;
        for(int i = 0; i < n; i++) 
        {
            cin>>a[i];
        }
        for(int i = 0; i < m; i++) 
        {
            cin>>b[i];
        }
        ll ans = 0;
        int j = 0, i = 0;
        
        for(; i < m; i++) 
        {   
            //cout<<ans<<' ';
            if(vis[b[i]]) ans++;
            else
            {
                for(; j < n; j++)
                {   
                    vis[a[j]] = 1;
                    if(a[j] == b[i])
                    {   
                        ans += ((j-i)*2+1);
                        break;
                    }
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

D. Santa's Bot

就是先选一个人x,然后从这个人的ki里面选一个y,然后随机分配给一个人z,如果z需要这个y的话,就是正确的操作,求正确操作的概率
说实话,英语TMD太重要了,我拿案例一来说吧
首先选择人1,如果选择2号礼物,只能分配给2,概率\(1/2 * 1/2 * 1/2 = 1/8\)
如果选择1号礼物,分配给1,2都可以,概率\(1/2 * 1/2 * 1 = 1/4\)
然后选择人2,只能选择1号礼物,分配给1,2都可以,概率\(1/2 * 1 * 1 = 1/2\)
所以总的概率\(7/8\)

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define cin(a) scanf("%d",&a)
#define pii pair<int,int>
#define ll long long
#define gcd __gcd
const int inf = 0x3f3f3f3f;
const int maxn = 1001000;
const int M = 998244353;

ll ksm(ll a,ll b)
{
    ll res = 1;
    while (b)
    {
        if(b&1) res = (res*a)%M;
        a = (a*a)%M;
        b /= 2;
    }
    return res%M;
}


vector<int> a[maxn];
int b[maxn];

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("data.in", "r", stdin);
    //freopen("data.out", "w", stdout);
#endif
    ll n,k;
    cin>>n;
    int sum = 0;
    for(int i = 0; i < n; i++) 
    {
        cin>>k;
        for(int j = 0,x; j < k; j++) 
        {
            cin>>x;
            sum++;
            a[i].push_back(x);
            b[x]++;
        }
    }
    ll nn = ksm(n,M-2);
    ll ans = 0;
    for(int i = 0; i < n; i++) 
    {
        ll res = 0;
        ll x = a[i].size();
        x = ksm(x,M-2);
        for(auto j : a[i])
        {
            res = (res + ((x*b[j])%M*nn)%M)%M;
        }
        ans = (ans+res)%M;
    }
    ans = (ans*nn)%M;
    cout<<ans<<endl;
    return 0;
}