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<bits/stdc++.h>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 secondint 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<bits/stdc++.h>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 secondint 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<bits/stdc++.h>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 secondint 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<bits/stdc++.h>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 secondint 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;
}

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

题目

A. Grass Field

B. Permutation

C. Schedule Management

D. Permutation Restoration

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