Description: \(t\)组询问,求第\(k\)个大于\(x\)且与\(p\)互质的数 Hint: \(x,k,p<=1e6,t<=30000\) Solution: 推出式子后,由于要求第k大,二分答案就好了 #include<bits/stdc++.h> using namespace std; const int mxn=1e6+5; int tot,vis[mxn],mu[mxn],sum[mxn],p[mxn]; void sieve(int lim) { mu…

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -. Example: Given a = 1 and b = 2, return 3. Credits:Special thanks to @fujiaozhu for adding this problem and creating all test cases. 这道题是CareerCup上的一道原题,难道…

Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT. 这道题让我们求两数相除,而且规定我们不能用乘法,除法和取余操作,那么我们还可以用另一神器位操作Bit Operation,思路是,如果被除数大于或等于除数,则进行如下循环,定义变量t等于除数,定义计数p,当t的两倍小于等于被除数时,进行如下循环,t扩大一倍,p扩大一倍,然后更…

How many integers can you find Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1796 Description   Now you get a number N, and a M-integers set, you should find out how many integers which are sm…

Divide two integers without using multiplication, division and mod operator. 不用乘.除.求余操作,返回两整数相除的结果,结果也是整数. 假设除数是2,相除的商就是被除数二进制表示向右移动一位. 假设被除数是a,除数是b,因为不知道a除以b的商,所以只能从b,2b,4b,8b.......这种序列一个个尝试 从a扣除那些尝试的值. 如果a大于序列的数,那么a扣除该值,并且最终结果是商加上对应的二进制位为1的数,然后尝试序…

原题链接在这里:https://leetcode.com/problems/sum-of-two-integers/ 题目: Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -. Example:Given a = 1 and b = 2, return 3. 题解: 两个数79和16 相加,不考虑进位相加得85, 只考虑进位进位是10, 85+10 = 95…

下面是今天写的几道题: 292. Nim Game You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take t…

A Simple Problem with Integers [题目链接]A Simple Problem with Integers [题目类型]线段树 成段增减+区间求和 &题解: 线段树 成段增减+区间求和 模板题 这种题真的应该理解并且可以流畅的独立码出来了 [时间复杂度]\(O(nlogn)\) &代码: #include <iostream> #include <cstdio> #include <cstring> using namespa…

Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -. Example:Given a = 1 and b = 2, return 3. 题目要求:计算两个整型的和,但是不能用+和- 我们知道a+b为((a&b)<<1)+(a^b),因此可以用递归的方法 class Solution { public: int getSum(int a, int b)…

题目 Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT 链接 https://leetcode.com/problems/divide-two-integers/ 答案 1.int的最大值MAX_INT为power(2,31)-1 = 2147483647 2.int的最小值MIN_INT为-power(2,31) = -21…