Educational Codeforces Round 100 (Rated for Div. 2) 虚拟参赛及补题。_codeforces虚拟赛加rate吗?

A题，很简单吧。可以说是。

• 思路
  9个一循环吧。

#include <bits/stdc++.h>

using namespace std;

const int N = 110;
typedef long long ll;
typedef pair<int, int> PII;


ll a[N];
int hsh[N];
int m;


void solve()
{
	int a,b,c;
	cin>>a>>b>>c;
	int res=a+b+c;
	if(res%9!=0)
	cout<<"NO"<<endl;
	else
	{
		int x=res/9;
		if(x<=a&&x<=b&&x<=c)
		cout<<"YES"<<endl;
		else
		{
			cout<<"NO"<<endl;
		}
	}

}
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		solve();
	}
}
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B题

• 题意
  

• 思路
  偶数，放1，级数放a【i】；或者奇数放1，偶数放a【i】；
  因为这样奇数或者偶数，为0.偶数或者奇数的差会小于s/2；
  但是不一定哪一个小于s/2，所以要测试一下，是奇数放1还是偶数放1.
  两个两个的看，无非有一个会小于两者之和。

#include <bits/stdc++.h>

using namespace std;

const int N = 55;
typedef long long ll;
typedef pair<int, int> PII;

ll a[N];
int hsh[N];
int m;

ll b[N];
void solve()
{
	int n;
	cin >> n;
	ll sum = 0;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}
	ll s = 0;
	for (int i = 1; i <= n; i++)
	{
		if (i % 2)
		{
			b[i] = 1;
			s +=(a[i]- b[i]);
		}
		else
		{
			b[i] = a[i];
			s += (a[i]-b[i]);
		}
	}
	
	if (s * 2 > sum)
	{
		for (int i = 1; i <= n; i++)
		{
			if (i % 2 == 0)
			{
				b[i] = 1;
			}
			else
			{
				b[i] = a[i];
			}
		}
	}
	for (int i = 1; i <= n; i++)
		cout << b[i] << " ";
	cout << endl;
}
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}
}

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D题

• 题意
  
• 思路

• 代码

#include <iostream>
#include <cstdio>

const int maxn=2e5+7;

using namespace std;

int T,n,sum,l,r;
int a[maxn];

int main()
{
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n);
		for (int i=1;i<=n;i++) scanf("%d",&a[i]);
		a[n+1]=2*n+1;
		l=0,r=0;
		sum=0;
		
		for (int i=1;i<=n;i++)
		{
			sum+=a[i]-a[i-1]-1;
			if (!sum) l++;
			     else sum--;
		}
		sum=0;
		for (int i=n;i>0;i--)
		{
			sum+=a[i+1]-a[i]-1;
			if (!sum) r++;
			     else sum--;
		}
		printf("%d\n",n-r-l+1);
	}
}

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C题

（模拟题）
题意

思路

代码

#include<bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
typedef long long ll;

ll T[N], X[N];

void solve()
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> T[i] >> X[i];
	ll nowx = 0, end = 0;
	ll res = 0;
	T[n + 1] = 2e10;
	for (int i = 1; i <= n; i++)
	{
		if (nowx == end) end = X[i];
		ll l = nowx, r;
		if (nowx > end)
		{
			if (nowx - end >= T[i + 1] - T[i])
			{
				nowx -= T[i + 1] - T[i];
			}
			else
				nowx = end;
		}
		else
		{
			if (end - nowx >= T[i + 1] - T[i])
				nowx += T[i + 1] - T[i];
			else
				nowx = end;
		}
		r = nowx;
		if (l > r)
			swap(l, r);
		if (l <= X[i] && X[i] <= r)
			res++;
	}
	cout << res << endl;
}

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}
}

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