2020牛客寒假算法基础集训营3

A

牛牛的DRB迷宫I

思路：dp求方案数

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 1000000007
const int maxn=1e5+100;
ll dp[55][55];
char s[55][55];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    dp[1][1]=1;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            cin>>s[i][j];
            if(s[i][j]=='R')
                dp[i][j+1]=(dp[i][j]+dp[i][j+1])%MOD;
            else if(s[i][j]=='D')
                dp[i+1][j]=(dp[i][j]+dp[i+1][j])%MOD;
            else if(s[i][j]=='B')
            {
                dp[i][j+1]=(dp[i][j]+dp[i][j+1])%MOD;
                dp[i+1][j]=(dp[i][j]+dp[i+1][j])%MOD;
            }
        }
    }
    cout<<dp[n][m]%MOD<<endl;
    return 0;
}

C

牛牛的数组越位

思路：模拟，不仅考虑<0还有>n，>m的情况，还有n*m与m*x+y的关系

#include<bits/stdc++.h>
using namespace std;
int a[1000010];
 
 
int main()
{
    int t;
    int n,m,p;
    int x,y,val;
    cin >>t;
    while(t--)
    {
        memset(a,0,sizeof(a));
        cin>>n>>m>>p;
        int len=n*m;
        int flag1=0;
        int flag2=0;
        while(p--)
        {
            
            cin>>x>>y>>val;
            int num=x*m+y;
            if(x<0||x>=n||y<0||y>=m)flag1=1;
             
            if(num<0||num>=len)
            {
                flag2=1;
               
            }
            else
            {
                a[num]=val;
            }
        }
            
            if(flag2)
            {
                cout<<"Runtime error"<<endl;
            }
            else
            {
                for(int i=0;i<len;i++)
                {
                    cout<<a[i]<<' ';
                    if((i+1)%m==0)cout<<endl;
                }
                if(flag1)
                {
                    cout<<"Undefined Behaviour"<<endl;
                }
                else
                {
                     cout<<"Accepted"<<endl;
                }
             
            }
          
    }
    return 0;
}

 

D	牛牛与二叉树的数组存储

思路：存位置即可，除去-1的情况

#include<iostream>
#include<cstdio>
#include<map>
using namespace std;
int a[100005];
int main()
{
    int n;
    scanf("%d",&n);
        map<int,int> m;
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&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;
}

F	牛牛的Link Power I

思路：刚开始想前缀和，算了一遍，没想到要两遍前缀和。。。注意模

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+105;
const int MOD=1e9+7;
long long a[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    string s;
    cin>>s;
    memset(a,0,sizeof(a));
    for(int i=0;i<n;i++)
    {
        if(s[i]=='1')
            a[i+1]++;
        if(i!=0)
        {
            a[i]+=a[i-1];         
            a[i]%=MOD;
        }
    }
    for(int i=0;i<n;i++)
        if(i!=0)
        {
            a[i]+=a[i-1];
            a[i]%=MOD;
        }
    long long sum=0;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='1')
        {
            sum+=a[i];
            sum%=MOD;
        }
    }
    printf("%lld\n",sum);
    return 0;
}

H

牛牛的k合因子数

思路：筛法求素数后应当就把1~n的合因子数存起来，否则会超时

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int pre[maxn];
const int MAX_N = 1e5+10;
int prime[MAX_N];
bool is_prime[MAX_N+1];
void Ai(){
    int p = 0;
    for(int i = 0; i <= maxn; i++) is_prime[i] = true;
    is_prime[0] = is_prime[1] =true;
    for(int i = 2; i <= maxn; i++){
        if(is_prime[i]){
            prime[p++] = i;
            for(int j = 2*i; j <= maxn; j+=i) is_prime[j] = false;
        }
    }
}
int n;
int res[maxn];
void he()
{
    for(int j=1;j<=n;j++)
    {
        int cnt=0;
        for(int i=1;i*i<=j;i++){
            if(j%i==0&&is_prime[i]==0)
                cnt++;
            if(j%i==0&&is_prime[j/i]==0&&i!=j/i)
                cnt++;
        }
        res[cnt]++;
    }
}
int main()
{
    Ai();
    int m;
    scanf("%d%d",&n,&m);
    he();
    int k;
    while(m--)
    {
        scanf("%d",&k);
        printf("%d\n",res[k]);
    }
    return 0;
}

I

牛牛的汉诺塔

思路：只会打表找规律。。。

#include<iostream>
using namespace std;
const int maxn = 7;
typedef long long ll;
ll a[maxn];
int main(){
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        if(i&1){
            a[2]=a[1]+a[4]+1;
            a[3]=a[4]+a[5];
            a[6]=a[3];
        }
        else{
            a[1] = a[2]+a[3];
            a[4] = a[1];
            a[5] = a[3]+a[6];
        }
    }
    cout<<"A->B:"<<a[1]<<endl;
    cout<<"A->C:"<<a[2]<<endl;
    cout<<"B->A:"<<a[3]<<endl;
    cout<<"B->C:"<<a[4]<<endl;
    cout<<"C->A:"<<a[5]<<endl;
    cout<<"C->B:"<<a[6]<<endl;
    cout<<"SUM:"<<(1ll<<n)-1<<endl;
    return 0;
}