Codeforces-Round#574 Div2

A题 Drinks Choosing

判断奇偶

#include<bits/stdc++.h>
using namespace std;
int arr[2000];
int main()
{
	int n,k;cin>>n>>k;int rem=0;
	for(int i=0;i<n;i++)
	{
		int num;
		cin>>num;arr[num]++;
	}
	for(int i=1;i<=k;i++)
	{
		if((arr[i]&1))
		{
			rem++;
		}
	}
	cout<<n-rem/2<<endl;
	return 0;
	
 } 

B题 Sport Mafia

水题

#include <iostream>
#include <cmath>
#define LL long long
using namespace std;
int main(){
    LL n,k;
    cin >> n >> k;
    LL tmp = 2*(k+n);
    LL maxn = 1ll*sqrt(tmp)+1;
    while(maxn){
        if(maxn*(maxn+3) == tmp){
            cout << n-maxn << endl;
            break;
        }
        maxn--;
    }
    return 0;
}

C题 Basketball Exercise

简单dp

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,h[2][200020],dp[2][200020];
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>h[0][i];
	for(int i=0;i<n;i++)
		cin>>h[1][i];
	dp[0][0]=h[0][0];dp[1][0]=h[1][0];
	for(int i=1;i<=n;i++)
	{
		dp[0][i]=max(h[0][i]+dp[1][i-1],dp[0][i-1]);
		dp[1][i]=max(h[1][i]+dp[0][i-1],dp[1][i-1]);
	}
	cout<<max(dp[0][n],dp[1][n])<<endl;
	return 0;
}

D1题 Submarine in the Rybinsk Sea (easy edition)

(f(a_1,a_2)+f(a_2,a_1)=f(a_1,a_1)+f(a_2,a_2)),即(sum=sum_{i=1}^n{f(a_i,a_i)}*n)

#include<bits/stdc++.h>
using namespace std;
typedef long long  ll;
ll sum=0;
ll pow(ll step)
{
	ll ans=1;
	while(step--)
	{
		ans=(ans*10)%998244353;
	}
	return ans;
}
ll get(ll num)
{
	ll ans=0,step=0;
	while(num>0)
	{
		step++;
		int d=num%10;
		num/=10;
		ans+=(d*(pow(step*2-1)%998244353)+d*(pow(step*2-2)%998244353))%998244353;
	}
	return ans;
}
int main()
{
	int n;
	cin>>n;int nn=n;
	while(n--)
	{
		ll num;cin>>num;
		sum+=get(num)%998244353;
		
	}
	sum=abs(sum);
	cout<<(sum*nn)%998244353<<endl;
	return 0;
}

D2题 Submarine in the Rybinsk Sea (hard edition)

按位运算，若12 和3 则，(ans=f(2,3)+1*2*10^2)

#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define FOR2(i,a,b) for(int i=a;i<=b;i++)
#define sync ios::sync_with_stdio(false);cin.tie(0) 
#define ll long long
using namespace std;
typedef struct{
	ll num;
	ll len;
}NUM; NUM nums[100010];
ll arr[100010];
ll p10[30];
ll mod=998244353 ;
ll ans=0;
ll getlen(ll a)
{
	int len=0;
	while(a>0)
	{
		a/=10;
		len++; 
		
	}
	arr[len]++;
	return len;
}
int main()
{
	sync;
	p10[1]=1;
	for(int i=2;i<30;i++)
	{
		p10[i]=(p10[i-1]*10)%mod;
	}
	int n;cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>nums[i].num;
		nums[i].len=getlen(nums[i].num);
	}
	for(int i=0;i<n;i++)
	{
		ll b=nums[i].num;
		int pos=0;
		while(b)
		{
			ll c=b%10;b/=10;pos++;
			for(int j=1;j<=10;j++)
			{//其他数的位数 
				if(j>=pos)
				{
					ans+=c*arr[j]*p10[pos*2];
					ans+=c*arr[j]*p10[pos*2-1];
				}
				else if(j<pos)
				{
					ans+=2*c*arr[j]*p10[pos+j];
				}
			}
			ans=ans%mod;
		}
	}
	cout<<ans%mod<<endl;
	return 0;
}