Summer training #4

D:找到两个数 一个是另一个的整数倍(1也算)

因为N是600000 调和级数为ln(n+1) 算一下 可以直接爆

#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-8
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
bool cmp(int x,int y){return x>y;}
int where[];
int visit[];
int main()
{
 //freopen("in.txt", "r", stdin);
 //freopen("out.txt", "w", stdout);
 int n,k;
 cin >> n >> k;
 int ans1=,ans2=;
 int t;
 for(int i=;i<=k;i++)
 {
        scanf("%d",&t);
        if(visit[t]!=) //==1
        {
        ans1=where[t];
        ans2=i;
        break;
        }
        visit[t]++;
        where[t]=i;
 }
 if(ans1!=&&ans2!=)
 {
        printf("%d %d\n",ans1,ans2);
        return ;
 }
 for(int i=;i<=n;i++)
        for(int j=*i;j<=n;j+=i)
 {
        if(visit[i]!=&&visit[j]!=)
        {
        ans1=where[i];
        ans2=where[j];
        break;
        }
 }
  if(ans1!=&&ans2!=)
 {
        printf("%d %d\n",ans1,ans2);
        return ;
 }
 printf("0 0\n");
  return ;
}

E:两个板之间的数之和是K的倍数?? 反正这么做过了

#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-8
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
bool cmp(int x,int y){return x>y;}
char a[];
//(a1+an)*n/2;
int main()
{
 //freopen("in.txt", "r", stdin);
 //freopen("out.txt", "w", stdout);
 int n,k;
 cin >> n;
 scanf("%s",a+);
 //printf("%s\n",a+1);
 int num=;
 int len=strlen(a+);
 //cout <<len<<endl;
 int first=,last=;
 int pop=;
 for(int i=;i<=len&&pop!=;i++)
 {
        if(a[i]=='')
        {
        //TS;
        for(int j=i;j<len;j+=n)
        {
        //cout << j<<endl;
        if(a[j+]=='')
        {
        first=i;
        last=j+;
        pop=;
        break;
        }
        }
        }
 }
 if(first!=&&last!=)
 {
        printf("%d %d",first,last);
        return ;
 }
 printf("0 0\n");
  return ;
}

G:水

H:2*3的砖铺 N*M的地板 以前CF好像做过 直接判min(n,m)>=2&&max(n,m)>=3  然后n*m%6==0

证明的话 引用别人的:

乘积是6的倍数那就说明有因子2和3：

1)一边是2的倍数恰好另一边是3的倍数，这种情况下显然可行；

2)一边是6的倍数另一边不是2也不是3的倍数，那另一边肯定是奇数，为1的时候不行，设边为N=2a+1=2(a-1)+3，那前面就以2x3的形式放最后一个以3x2的形式放，由于一边为6的倍数那么就是一定可以放得下的

#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-8
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
int main()
{
 //freopen("in.txt", "r", stdin);
 //freopen("out.txt", "w", stdout);
 int k;
 cin >> k;
 while(k--)
 {
        ll n,m;
        cin >> n >> m;
        ll sum=n*m;
        if(min(n,m)<||max(n,m)<)
        {
        cout<<"No"<<endl;
        continue;
        }
        if(sum%==)
        cout <<"Yes"<<endl;
        else
        cout <<"No"<<endl;
 }
  return ;
}

I:阅读理解 把最长的R B先连起来

#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-8
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
int red[],blue[];
bool cmp(int x,int y){return x>y;}
int main()
{
 //freopen("in.txt", "r", stdin);
 //freopen("out.txt", "w", stdout);
 int n;
 while(scanf("%d",&n)==)
 {
        mem(red,);
        mem(blue,);
        int pop=;
        int pop1=;
        int number=;
        int ans=;
        char ch;
        int  j;
        for(int i=;i<=n;i++)
        {
        cin >> j>>ch;
        if(ch=='R')
        {
        red[++pop]=j;
        }
        else
        {
        blue[++pop1]=j;
        }
        }
        sort(red+,red+pop+,cmp);
        sort(blue+,blue+pop1+,cmp);
        for(int i=;i<=min(pop,pop1);i++)
        {
        ans+=red[i]+blue[i]-;
        }
        cout <<ans<<endl;
 }
  return ;
}

J:按条件读取字符 组成词组 输出

#include <bits/stdc++.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#define foror(i,a,b) for(i=a;i<b;i++)
#define foror2(i,a,b) for(i=a;i>b;i--)
#define EPS 1.0e-8
#define PI acos(-1.0)
#define INF 3000000000
#define MOD 1000000009
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
//using ll = long long;
//using ull= unsigned long long;
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que;
typedef long long ll;
//stack<string> a;
int pop=;
map<string,int> mp;
string strans;
int ans=;
char c[];
//(a1+an)*n/2;
void print()
{
for(map<string,int>::iterator it=mp.begin();it!=mp.end();it++)
{
        if(it->second>ans)
        {
        strans=it->first;
        ans=it->second;
        }
}
cout << strans<<":"<<ans<<endl;
}
int main()
{
 //freopen("in.txt", "r", stdin);
 //freopen("out.txt", "w", stdout);
  while(gets(c))
  {
        int len=strlen(c);
        if(len==&&c[]=='#')
        {
        print();
        mp.clear();
        strans="";
        ans=;
        continue;
        }
        else
        {
        for(int i=;i<len;i++)
        {
                string start="";
                if(c[i]>='a'&&c[i]<='z')
                {
                int cur=i;
                while(c[cur]>='a'&&c[cur]<='z')
                {
                start+=c[cur];
                cur++;
                }
                i=cur;
                if(c[cur]==','||c[cur]=='\0')
                continue;
                start+=' ';
                while(c[cur]==' ')
                cur++;
                while(c[cur]>='a'&&c[cur]<='z')
                {
                start+=c[cur];
                cur++;
                }
                //cout <<"start="<<start<<" ";
                //cout <<endl;
                mp[start]++;
                }
        }
        }
  }
  return ;
}