Turn from: http://blog.csdn.net/newhappy2008/article/details/6857098
Question: There are n+2 number, n number appeared even several times, 2 numbers appear odd several times (these two numbers are not equal), ask the space complexity of O (1), find out these two numbers, do not need to know the exact location, just need to know these two values.Solution: If only
Title Link: http://lightoj.com/volume_showproblem.php?problem=1278Test instructions: give you a number n (nFor example: 15 = 7+8 = 4+5+6 = 1+2+3+4+5, so 15 corresponding answer is 3, there are three kinds;We are now equivalent to known arithmetic progression and sum = N, the other first item is A1, Total m, then am = a1+m-1;sum = m* (a1+a1+m-1)/2-----> a1 = sum/m-(m-1)/2A1 and M must be integers, so sum%m = 0 and (m-1)%2=0, so M is the factor of sum,
The sword refers to the offer topic: http://ac.jobdu.com/problem.php?pid=1516
Title Description:
Enter an array of integers to implement a function that adjusts the order of the numbers in the array so that all the odd digits are placed in the first half of the array, all the even digits are located in the second half of the array, and the relative positions between the odd and
Problem description
Give you n integers and calculate the product of all the odd numbers in them.
Input
The input data contains multiple test instances. Each test instance occupies one row. The first number of each row is N, indicating that there are n data in this group, followed by N integers, you can assume that each group of data must have at least one odd
To find the product of odd numbersTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 56573 Accepted Submission (s): 36490Problem Descriptiongive you n integers and ask for the product of all the odd numbers in them.InputThe input data contains multiple test instances, one row per test instance, and the first number
Topic Link: PortalTest instructions: Divides the given n into successive numbers and has at least two numbers to see how many options are available.Analysis:A + (A + 1) + (A + 2) + ... +(A + k-1) = N; ===> (2*a + k-1) * k = 2*n;===> (2*a-1) *k = 2*n-k*k;===> 2*a-1 = 2*n/k-K;The left side of the equation is odd, then the right side must be odd, then K must be an odd
How to count the number of odd elements in a JS array?This is a question raised by the group of friends, most of the group of friends give the traversal and then the 2 modulo, the final result.This kind of writing is the most easy to get, then there are no other ideas?Here I provide another way of thinking, we know that the odd
Title: Enter an array of shapes, reorder the arrays so that all the odd numbers are in the front even, and the relative positions between the odd and the even are not changed.Thought: Iterates through all arrays starting at the beginning of the array. When an even number is encountered, the even is packaged, that is, the count of the even numbers so far, and thes
Title: Enter an array of integers to implement a function to adjust the order of the numbers in the array, so that all the odd digits are in the first half of the array, all the even digits are located in the second half of the array, and the relative positions of the odd and odd, even and even are guaranteed.Idea: Build an array, sweep both sides, first fill in
PHP returns an odd and even number of php statements every Saturday.
Requirement description: determines whether a week is a double week or a single week based on the Saturday date. that is, if the date on Saturday is a singular number, this week is a single break. if it is a double number, it is a double brea
Odd-order Rubik's CubeTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 3071 Accepted Submission (s): 1614Problem description The elements of an n-order phalanx are,..., n^2, each of its rows, each column and 2 diagonal elements and equal, soSquare is called Rubik's Cube. n is an odd number when we have 1 constr
Title: Enter an array of integers to implement a function to adjust the order of the numbers in the array,So that all the odd digits are in the first half of the array, and all the even digits are in the second half of the array.This question is actually very simple, of course, without considering the efficiency of the situation can be consideredWe'll start the array from the previous index as long as the even num
Evaluate a regular expression that determines an odd number between 1 and 99 and uses a regular expression to determine whether the input number is an odd number between 1 and 99, including 1 and 99. thank you. ------ Solution -------------------- This post was last edited b
Title: Enter an array of integers to implement a function to adjust the order of the numbers in the array, so that all the odd digits are in the first half of the array, all the even digits are in the second half of the array, and the relative positions of the odd and odd, even and even are constant. Public voidReorderarray (int[] Array) { for(inti = 0; i ) {
The sword refers to the offer topic: http://ac.jobdu.com/problem.php?pid=1516
title Description:
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Enter an array of integers, Implement a function to adjust the order of the numbers in the array. Causes all the odd digits to be in the first half of the array. All the even digits are located in the second half of the array and are guaranteed to be odd
Odd-even number Time
limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total Submission (s): 943 Accepted Submission (s): 504
problem Description for
a number,if the length of continuous o DD digits is even and the length of continuous even digits are odd,we call it
. Notice:for each integer in the input n (nOutputOutput N^n ' s digital root on a separate line of the output.Sample Input240Sample Output44AuthorEddyThe main topic: to give you a positive integer n, the number of N to add up, assuming that the result is less than 10, the result is N of the number of the root, assuming that more than 10, then the upper results of the figures on each of you add up. Now give
Test instructions: Xiao Ming to buy three houses, the three houses constitute a triangle, known as the coordinates of the house N, any three houses are not in a straight line, but also known as the coordinates of the M treasure, asked the house composed of triangles within the triangle has an odd number of the triangle there are many. Data range: N (3~100), M (1~1000)Analysis:Simple computational geometry.
http://poj.org/problem?id=1284Test instructions: Given an odd prime p, the number of the original root of P is obtained.Original root: {(xi mod p) | 1 Conclusion: The number of the original root of the singular prime P is phi (P-1).Prove:
for the given prime number p, the first thing to be clear: the Motone of P mu
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