2020牛客寒假算法基础集训营3

2020牛客寒假算法基础集训营3

A

#include<bits/stdc++.h>
using namespace std;
const int MAXN=55;
const long long mod=1e9+7;
int n,m;
char s[MAXN][MAXN];
long long dp[MAXN][MAXN];
long long dp_dfs(int x,int y)
{
    if(x>n||y>m)return 0;
    if(dp[x][y]!=-1)return dp[x][y];
    if(s[x][y]=='D')return dp[x][y]=dp_dfs(x+1,y);
    if(s[x][y]=='R')return dp[x][y]=dp_dfs(x,y+1);
    if(s[x][y]=='B')return dp[x][y]=(dp_dfs(x+1,y)+dp_dfs(x,y+1))%mod;
}
int main()
{
    scanf("%d %d",&n,&m);
    memset(dp,-1,sizeof(dp));
    dp[n][m]=1;
    for(int i=1;i<=n;++i)
    {
        scanf("%s",s[i]+1);
    }
    printf("%lld\n",dp_dfs(1,1));
    return 0;
}

B

C

#include<bits/stdc++.h>
using namespace std;
const int N = 1005;
typedef long long ll;
int a[N][N];
int main()
{
    ll t;cin >> t;
    while (t--)
    {
        memset(a, 0, sizeof(a));
        ll flag = 1;
        ll n, m, p;cin >> n >> m >> p;
        for (int i = 0;i < p;++i)
        {
            ll x, y, val;cin >> x >> y >> val;
            const int sum = m * x + y;
            if (sum >= n * m || sum < 0)//Re
                flag = 2;
            else if (flag==1&&(x < 0 || y<0 || y>=m||x>=n))//UD
                flag = 3;
            if(flag!=2)a[sum/m][sum%m] = val;
        }
        if (flag == 2)
            puts("Runtime error");
        else
        {
            for (int i = 0;i < n;++i)
            {
                cout << a[i][0];
                for (int j = 1;j < m;++j)
                {
                        cout << " " << a[i][j];
                }
                cout << endl;
            }
            if (flag == 1)puts("Accepted");
            else puts("Undefined Behaviour");
        }
    }
    return 0;
}

D

/*************************************************************************
	> File Name: d.cpp
	> Author: 
	> Mail: 
	> Created Time: 六  2/ 8 20:19:27 2020
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long ll;

vector<int> a(1e5 + 5);

int main(int argc, char *argv[]) {
    SIS;
    int n;
    while(cin >> n){
        map<int,int> m;
        int cnt =  0;
        for(int i = 1;i <= n;i++){
            cin >> a[i];
            if(a[i] > 0){
                cnt++;
                m[a[i]] = i;
            }
        }
        printf("The size of the tree is %d\n",cnt);
        printf("Node %d is the root node of the tree\n",a[1]);
        for(int i = 1;i <= cnt ;i ++){
            printf("The father of node %d is %d,",i,m[i] / 2 == 0 ? - 1 : a[m[i]/2]);
            printf(" the left child is %d, and the right child is %d\n",m[i] * 2 > n ? -1:a[m[i] * 2],m[i] * 2 + 1 > n ? -1 : a[m[i] * 2 + 1]);
        }
    }
    return 0;
}

H

使用埃式筛,将每一个质数筛出,如果x是合数,则将1~n范围内能被该合数x 筛出的合数 ++,因为我们知道一个合数有多少个合数因子就能被筛几次

然后桶排序即可

/*************************************************************************
	> File Name: h.cpp
	> Author: 
	> Mail: 
	> Created Time: 一  2/10 10:48:26 2020
 ************************************************************************/

#include<bits/stdc++.h>
using namespace std;
#define SIS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
typedef long long ll;
const int maxn = 1e5 + 5;

int f[maxn];
int cnt[maxn];
int prime[maxn];
void init(int n){
    for(int i = 2;i <= n;i++){
        prime[i] = true;
    }
    for(int i = 2;i <= n;i++){
        if(prime[i]){
            for(int j = i + i; j <= n; j += i){
                prime[j] =false;
            }
        }else{
            for(int j = i; j <= n; j += i){
                cnt[j]++; // 代表被筛查了几次,一个数有几个合数就能被筛查多少次
            }
        }
        f[cnt[i]]++; // 桶排序,出现过1次的++
    }
}


int main(int argc, char *argv[]) {
    SIS;
    int n;
    int k;
    cin >> n >> k;
    init(n);
    for(int i = 0;i  < k;i ++){
        int x;
        cin >> x;
        cout << f[x] << endl;
    }
    return 0;
}