2020杭电多校第十场题解3、4、11

发布于 2020-08-24  0 次阅读


Mine Sweeper(构造)

思路:

官方题解给的清清楚楚明明白白

在这里插入图片描述

AC Code:

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3f
#define int long long
// #define TDS_ACM_LOCAL
const int N=2e5 +9;
int n;
void solve(){
    cin>>n;
    if(n<24){
        cout<<"1 "<<n+1<<endl;
        for(int i=1; i<=n+1; i++){
            if(i&1) cout<<".";
            else    cout<<'X';
        }
        cout<<endl;
    }
    else{
        cout<<"25 25"<<endl;
        int t1=n/8, t2=0;
        while(1){
            if((n-t1*8)%3==0)   {t2=(n-t1*8)/3; break;}
            t1--;
        }
        for(int i=1; i<25; i++){
            if(i&1) for(int j=1; j<=25; j++)    cout<<".";
            else{
                for(int j=1; j<=25; j++){
                    if(!(j&1) && t1>0)  cout<<"X", t1--;
                    else                cout<<".";
                }
            }
            cout<<endl;
        }
        for(int k=1; k<=25-t2; k++) cout<<".";
        for(int k=25-t2+1; k<=25; k++)  cout<<"X";
        cout<<endl;
    }
    return ;
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
#ifdef TDS_ACM_LOCAL
    freopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);
    freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);
#endif
    int T;
    cin>>T;
    while(T--)  solve();
    return 0;
}

Permutation Counting

题目大意:

对于一个字符串a,可以形成一个字符串b,方法为

现在我们给您一个整数n和一个长度为n-1的序列b
求序列等于b的长度为n的排列数

思路:

考虑d p [ i ] [ j ] ,其中i ii表示当前的位置,j 为当前最后一位的数字的大小
在b中如果下一个为1,说明后面一个数比前一个数小,即为从1~j-1这些数作为末尾的情况转移过来。
记得前缀和优化一下

AC Code:

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3f
#define int long long
// #define TDS_ACM_LOCAL
const int mod=1e9 +7;
const int N=5e3 +9;
int n, a[N], dp[N][N];
void solve(){
    cin>>n;
    memset(dp,0,sizeof(dp));
    dp[1][1]=1;
    for(int i=1; i<n; i++) cin>>a[i];
    int sum=0;
    for(int i=2; i<=n; i++){
        sum=0;
        if(a[i-1]){
            for(int j=i-1; j; j--)
                sum=(sum+dp[i-1][j])%mod, dp[i][j]=(dp[i][j]+sum)%mod;  
        }
        else{
            for(int j=2; j<=i; j++)
                sum=(sum+dp[i-1][j-1])%mod, dp[i][j]=(dp[i][j]+sum)%mod;
        }
    }
    int ans=0;
    for(int i=1; i<=n; i++) ans=(ans+dp[n][i])%mod;
    cout<<ans<<endl;
    return ;
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
#ifdef TDS_ACM_LOCAL
    freopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);
    freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);
#endif
    int T;
    cin>>T;
    while(T--)  solve();
    return 0;
}

Task Scheduler

思路:

参考大佬的思路吧:大佬

AC Code:

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
#define endl '\n'
#define INF 0x3f3f3f3f
#define int long long
// #define TDS_ACM_LOCAL
const int N=1e4 +9;
int n, m, k;

struct node{
    int v, idx;
    bool operator< (const node& a)const {
        if(v!=a.v)  return v>a.v;
        return idx<a.idx;
    }
}a[N];
void solve(){
    cin>>n>>m>>k;
    for(int i=1; i<=n; i++){
        cin>>a[i].v;
        a[i].idx=i;
    }
    if(k!=0)    sort(a+1,a+1+n);
    for(int i=1; i<n; i++)  cout<<a[i].idx<<" ";
    cout<<a[n].idx<<endl;
    return ;
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
#ifdef TDS_ACM_LOCAL
    freopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);
    freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);
#endif
    int T;
    cin>>T;
    while(T--)  solve();
    return 0;
}

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