Codeforces Round #276 (Div. 2)

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本文深入探讨了信息技术领域的核心内容，从基础到高级，包括前端开发、后端开发、移动开发等细分技术领域，提供了关键信息和技术洞察。

One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were x details in the factory storage, then by the end of the day the factory has to produce $\text{[math]}$ (remainder after dividing x by m) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory.

The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by m).

Given the number of details a on the first day and number m check if the production stops at some moment.

Input

The first line contains two integers a and m (1 ≤ a, m ≤ 10⁵).

Output

Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".

简单题

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
  
using namespace std;

int main()
{
	int a, m;
	while (~scanf("%d%d", &a, &m))
	{
		map<int, bool> qu;
		qu.clear();
		bool flag = false;
		while(1)
		{
			a = a + a % m;
			a %= m;
			if (a == 0)
			{
				flag = true;
				break;
			}
			if (qu[a])
			{
				break;
			}
			else
			{
				qu[a] = 1;
			}
		}
		if (flag)
		{
			printf("Yes\n");
		}
		else
		{
			printf("No\n");
		}
	}
	return 0;
}

Many computer strategy games require building cities, recruiting army, conquering tribes, collecting resources. Sometimes it leads to interesting problems.

Let's suppose that your task is to build a square city. The world map uses the Cartesian coordinates. The sides of the city should be parallel to coordinate axes. The map contains mines with valuable resources, located at some points with integer coordinates. The sizes of mines are relatively small, i.e. they can be treated as points. The city should be built in such a way that all the mines are inside or on the border of the city square.

Building a city takes large amount of money depending on the size of the city, so you have to build the city with the minimum area. Given the positions of the mines find the minimum possible area of the city.

Input

The first line of the input contains number n — the number of mines on the map (2 ≤ n ≤ 1000). Each of the next n lines contains a pair of integers x_i and y_i — the coordinates of the corresponding mine ( - 10⁹ ≤ x_i, y_i ≤ 10⁹). All points are pairwise distinct.

Output

Print the minimum area of the city that can cover all the mines with valuable resources.

Sample test(s)

Input

2
0 0
2 2

Output

Input

2
0 0
0 3

Output

简单题

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
  
using namespace std;

struct node
{
	int x, y;
}pp[1010];

int cmp1(node a, node b)
{
	return a.x < b.x;
}

int cmp2(node a, node b)
{
	return a.y < b.y;
}

int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		for (int i = 0; i < n; ++i)
		{
			scanf("%d%d", &pp[i].x, &pp[i].y);
		}
		int l, r, u, d;
		sort(pp, pp + n, cmp1);
		l = pp[0].x;
		r = pp[n - 1].x;
		sort(pp, pp + n, cmp2);
		d = pp[0].y;
		u = pp[n - 1].y;
		__int64 a = (__int64)max(r - l, u - d);
		printf("%I64d\n", a * a );
	}
	return 0;
}

Let's denote as $\text{[math]}$ the number of bits set ('1' bits) in the binary representation of the non-negative integer x.

You are given multiple queries consisting of pairs of integers l and r. For each query, find the x, such that l ≤ x ≤ r, and $\text{[math]}$ is maximum possible. If there are multiple such numbers find the smallest of them.

Input

The first line contains integer n — the number of queries (1 ≤ n ≤ 10000).

Each of the following n lines contain two integers l_i, r_i — the arguments for the corresponding query (0 ≤ l_i ≤ r_i ≤ 10¹⁸).

Output

For each query print the answer in a separate line.

Sample test(s)

Input

Output

1
3
7

我想到的思路是，先判断区间[l，r]之间有没有2^n - 1这样的数，如果有，那么答案就是这个数；否则，找到两个这样的数A, B，包含这个区间，然后相加除以2得到C，然后判断与区间的关系，如果在区间左边，就用C + B / 2,否则就用A + C / 2，直到在区间内，可惜超时了，贴下代码

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
  
using namespace std;

__int64 l, r;
__int64 a[70];

int main()
{
	int n;
	__int64 li, ri, mid, lii, rii;
	a[0] = 1;
	for (int i = 1; i <= 59; ++i)
	{
		a[i] = a[i - 1] << 1;
		a[i]++;
	}
	scanf("%d", &n);
	while(n--)
	{
		scanf("%I64d%I64d", &l, &r);
		if (l == r)
		{
			printf("%I64d\n", r);
			continue;
		}
		bool flag = false;
		for (int i = 59; i >= 0; --i)
		{
			if (a[i] >= l && a[i] <= r)
			{
				printf("%I64d\n", a[i]);
				flag = true;
				break;
			}
		}
		if (flag)
		{
			continue;
		}
		for (int i = 0; i <= 59; ++i)
		{
			if (a[i] > r)
			{
				ri = rii = a[i];
				break;
			}
		}
		for (int i = 59; i >= 0; --i)
		{
			if (a[i] < l)
			{
				li = lii = a[i];
				break;
			}
		}
		while (1)
		{
			mid = (li + ri) >> 1;
			if (mid < l)
			{
				li = mid;
				ri = rii;
			}
			else if(mid > r)
			{
				ri = mid;
				li = lii;
			}
			else
			{
				printf("%I64d\n", mid);
				break;
			}
		}
	}
	return 0;
}