AtCoder Beginner Contest 174

cf这次要隔好久才能打，来打场abc放松下，结果赛中c做不出来了哈哈哈哈哈哈也是没谁了，赛后发现其实还是挺巧妙的??

A - Air Conditioner

判断是否小于30

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstdlib>
#include<string>
#include<unordered_map>
#include<vector>
#include<list>
#include<map>
#include<set>
#define fi first
#define se second
#define pb push_back
#define me(a,b) memset(a,b,sizeof(a))
#define INIT() std::ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int MAX_N=200000+5;
const int MAX_M=100000+5;
const int INF=0x7fffffff;
const int inf=1000000000;
const double EPS=1e-6;
const ull base=123;
const ll mod=1e9+7;
const double pi=4*atan(1.0);
int main()
{
   
    int n;
    cin>>n;
    if(n>=30)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
}

B - Distance

距离公式判断是否超过d

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstdlib>
#include<string>
#include<unordered_map>
#include<vector>
#include<list>
#include<map>
#include<set>
#define fi first
#define se second
#define pb push_back
#define me(a,b) memset(a,b,sizeof(a))
#define INIT() std::ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int MAX_N=200000+5;
const int MAX_M=100000+5;
const int INF=0x7fffffff;
const int inf=1000000000;
const double EPS=1e-6;
const ull base=123;
const ll mod=1e9+7;
const double pi=4*atan(1.0);
int main()
{
   
    int n;
    double d;
    scanf("%d%lf",&n,&d);
    int ans=0;
    while(n--)
    {
   
        double x,y;
        scanf("%lf%lf",&x,&y);
        if((x*x+y*y)<=d*d)
            ans++;
    }
    cout<<ans<<endl;
}

C - Repsept

赛中居然卡了，主要看到最后那个样例人就傻了，完全忘记了取模这个东西，直接看代码吧，还挺直观的

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstdlib>
#include<string>
#include<unordered_map>
#include<vector>
#include<list>
#include<map>
#include<set>
#define fi first
#define se second
#define pb push_back
#define me(a,b) memset(a,b,sizeof(a))
#define INIT() std::ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int MAX_N=200000+5;
const int MAX_M=100000+5;
const int INF=0x7fffffff;
const int inf=1000000000;
const double EPS=1e-6;
const ull base=123;
const ll mod=1e9+7;
const double pi=4*atan(1.0);
int main()
{
   
    int n;
    cin>>n;
    int k=7;
    int x=k%n;
    int ans=1;
    while(x)
    {
   
        if(ans>1000000)
            break;
        x=x*10+7;
        x%=n;
        ans++;
    }
    if(!x)
    {
   
        cout<<ans<<endl;
    }
    else
        cout<<-1<<endl;
}

D - Alter Altar

要求没有W在R的左边，可以互换也可以直接变颜色，但显然互换在任何情况下都更优，所以直接双指针贪心的互换就行。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstdlib>
#include<string>
#include<unordered_map>
#include<vector>
#include<list>
#include<map>
#include<set>
#define fi first
#define se second
#define pb push_back
#define me(a,b) memset(a,b,sizeof(a))
#define INIT() std::ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int MAX_N=200000+5;
const int MAX_M=100000+5;
const int INF=0x7fffffff;
const int inf=1000000000;
const double EPS=1e-6;
const ull base=123;
const ll mod=1e9+7;
const double pi=4*atan(1.0);
char s[MAX_N];
int main()
{
   
    int n;
    cin>>n;
    scanf("%s",s);
    int i,j;
    int len=strlen(s);
    i=0;
    j=len-1;
    int ans=0;
    for(i=0; i<j; i++)
    {
   
        if(s[i]=='W')
        {
   
            while(j>i&&s[j]=='W')
            {
   
                j--;
            }
            if(j!=i)
            {
   
                s[j]='W';
                ans++;
            }
        }
    }
    cout<<ans<<endl;
}

E - Logs

n根棍子，总共可以切k次，要求切完后最长的那根木棍的最小值。
对于一个长len的木棍，假如我想切成最大不超过x长的每段，显然要切len/x-len%x==0.所以这题实际只要二分切出的最长段，然后通过总共要切多少次来判断就行，注意二分的过程中除数为0的情况，我赛中因为这re了一发。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstdlib>
#include<string>
#include<unordered_map>
#include<vector>
#include<list>
#include<map>
#include<set>
#define fi first
#define se second
#define pb push_back
#define me(a,b) memset(a,b,sizeof(a))
#define INIT() std::ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int MAX_N=200000+5;
const int MAX_M=100000+5;
const int INF=0x7fffffff;
const int inf=1000000000;
const double EPS=1e-6;
const ull base=123;
const ll mod=1e9+7;
const double pi=4*atan(1.0);
int a[MAX_N];
int n,k;
int check(int x)
{
   
    int ans=0;
    int i;
    for(i=1; i<=n; i++)
    {
   
        if(a[i]>x)
        {
   
            ans+=a[i]/x;
            if(a[i]%x==0)
                ans--;
        }
        if(ans>k)
            return 1;
    }
    return 0;
}
int main()
{
   
    cin>>n>>k;
    int i;
    for(i=1; i<=n; i++)
        scanf("%d",&a[i]);
    int l=0;
    int r=1000000000;
    int mid=(l+r)/2;
    while(l<=r)
    {
   
        if(mid==0)
            break;
        if(check(mid))
        {
   
            l=mid+1;
        }
        else
            r=mid-1;
        mid=(l+r)/2;
    }
    cout<<mid+1<<endl;
}