Educational Codeforces Round 131 (Rated for Div. 2) A-D题解

A. Grass Field

题解：
（1）若四个位置全是1，则输出2
（2）若全为0则输出0，
（3）其他情况都输出1

/*
 author : Mzx
*/
#include

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> PII;
const ll mod=1e9+7;
const int INF=0x3f3f3f3f;
const int maxn=1e5+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define x first
#define y second


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

void solve()
{
    int a,b,c,d;
    cin>>a>>b>>c>>d;
    if(a==0&&b==0&&c==0&&d==0) cout<<0<<endl;
    else if(a==1&&b==1&&c==1&&d==1) cout<<2<<endl;
    else cout<<1<<endl;
}

int main()
{
    ios;
    int t;cin>>t;
    while(t--) {
        solve();
    }
    return 0;
}

B. Permutation

题解：一道思维题，很容易得知d=2时花费总是最大的，证明比较容易就不在此赘述。

/*
 author : Mzx
*/
#include

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> PII;
const ll mod=1e9+7;
const int INF=0x3f3f3f3f;
const int maxn = 2e5+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define x first
#define y second


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

void solve() {
    int n;cin>>n;
    cout<<2<<endl;
    for(int i = 1;i <= n;i += 2)
    {
        int p = i;
        while(p <= n)
        {
            cout<<p<<' ';
            p *= 2;
        }
    }
    cout<<endl;
}


int main()
{
    ios;
    int t; cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}

C. Schedule Management

题解：二分答案，check函数里对n个人分情况算
（1）如果s[i] >= mid的话，用need计数，计算剩下多少需要去完成的任务，
（2）如果s[i] < mid的话，用sum记录剩下的时间可以完成多少“不擅长的任务”.
最后比较一下need和sum的大小即可。

/*
 author : Mzx
*/
#include

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> PII;
const ll mod=1e9+7;
const int INF=0x3f3f3f3f;
const int maxn = 2e5+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define x first
#define y second


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

ll s[maxn];
ll n,m;
bool check(ll x)
{
    ll need = 0;
    ll sum = 0;
    for(int i = 1;i <= n;i++)
    {
        if(s[i] >= x) need += (s[i] - x);
        else sum += (x-s[i])/2;
    }
    if(need <= sum) return true;
    else return false;
}

void solve() {
    cin>>n>>m;
    memset(s,0,sizeof s);
    for(int i = 1;i<=m;i++)
    {
        int x;
        cin>>x;
        s[x]++;
    }
    ll l = 1,r = INF;
    while(l < r)
    {
        ll mid = (l + r)>>1;
        if(check(mid)) r = mid;
        else l = mid + 1;
    }
    cout<<l<<endl;
}

int main()
{
    ios;
    int t; cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}

D. Permutation Restoration

题解：b[i]通过转换可以得到不等式：

就可以转换成一个基本的区间贪心模型了。
PS：需要注意的是，要特判b[i] = 0的情况。

/*
 author : Mzx
*/
#include

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int,int> PII;
const ll mod=1e9+7;
const int INF=0x3f3f3f3f;
const int maxn = 2e5+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define x first
#define y second


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

void solve() {
    int n;
    cin >> n;
    vector<int> b(n);
    for (int j = 0; j < n; j++){
        cin >> b[j];
    }
    vector<int> L(n), R(n);
    for (int j = 0; j < n; j++){
        if (b[j] == 0){
            L[j] = j + 2;
            R[j] = n;
        } else {
            L[j] = (j + 1) / (b[j] + 1) + 1;
            R[j] = (j + 1) / b[j];
        }
    }
    vector<vector<int>> P(n+1);
    for (int j = 0; j < n; j++){
        P[L[j]].push_back(j);
    }
    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
    vector<int> a(n);
    for (int j = 1; j <= n; j++){
        for (int x : P[j]){
            pq.push(make_pair(R[x], x));
        }
        int p = pq.top().second;
        pq.pop();
        a[p] = j;
    }
    for (int j = 0; j < n; j++){
        cout << a[j]<<' ';
    }
    cout<<endl;
}


int main()
{
    ios;
    int t; cin>>t;
    while(t--)
    {
        solve();
    }
    return 0;
}