Codeforces Round #698 (Div. 2) 题解_codeforces #698 div2

Codeforces Round #698 (Div. 2)

A. Nezzar and Colorful Balls

找数字出现次数最多的个数

map<int,int> book;
int a[maxn];
int main()
{
	int t;
	cin >> t;
	while (t--) {
		book.clear();
		int n;
		cin >> n;
		for(int i=1;i<=n;i++)
		{
			cin >> a[i];
			book[a[i]]++;
		}
		int maxx=-1;
		for(int i=1;i<=n;i++)
		{
			maxx=max(maxx,book[a[i]]);
		}
		cout << maxx << endl;
	}
}
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B. Nezzar and Lucky Number

构造，分情况讨论

1. $a_i\ge 10\ast d$，那么一定可以构造成(10d+$a_i$%d)+kd的形式（k为整数）
2. $a_i<10\ast d$，判断能否构造为10n+kd的形式（k为正整数且k<10）

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int main()
{
	ios;
	int t;
	cin >> t;
	while (t--) {
		int q,d;
		cin >> q >> d;
		for(int i=1;i<=q;i++)
		{
			int num;
			cin >> num;
			bool flag=false;
			if(num>=10*d) { cout<< "YES" << endl; continue;}
			for(int k=1;k<=9;k++)
			{
				if(num<k*d) break;
				if((num-k*d)%10==0) flag=true;
			}
			if(flag) cout << "YES" << endl;
			else cout << "NO" << endl;
		}
	}
}
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---

C. Nezzar and Symmetric Array

ll d[maxn],dd[maxn];
map<ll,int> book;
int main()
{
	int t;
	scanf("%d",&t);
	while (t--) {
		book.clear();
		ll n,cnt=0;
		scanf("%lld",&n);
		for(int i=1;i<=2*n;i++)
		{
			scanf("%lld",&d[i]);
			book[d[i]]++;
			if(book[d[i]]==1) dd[++cnt]=d[i];
		}
		sort(dd+1, dd+1+cnt);
		bool flag=true;
		for(int i=1;i<=2*n;i++)
		{
			if(book[d[i]]!=2)
			{
				flag=false;
				break;
			}
		}
		ll a1r=0;
		if(flag)
		{
			for(int i=n;i>=2;i--)
			{
				ll tmp=dd[i]-dd[i-1],tt=2*(i-1);
				if(tmp%tt) { flag=0; break;}
				a1r+=(n-i+1)*tmp/tt;
			}
		}
		if(dd[1]<=2*a1r || (dd[1]-2*a1r)%n || (dd[1]-2*a1r)/n%2) flag=0;
		if(flag) printf("YES\n");
		else printf("NO\n");
	}
}
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