From stalking to plans of settling in Kathmandu — Mastermind behind the abduction of Snapdeal’s Dipti Sarna reveals chilling details.

## Sony’s latest 4K HDR TVs are now up for preorder

Sony’s 2016 lineup of TVs can be preordered beginning today, and they’ll be arriving at retailers like Amazon and Best Buy beginning next month. We took a look at the sets back in January at CES; this year, Sony is putting emphasis on an improved HDR picture and the ultra-thin bezels in its top models. The buzzwords haven’t changed much, with the company talking up its proprietary technologies like X-tended Dynamic Range PRO and Triluminos Display, which has been “further enhanced for color accuracy” this year. Sony concedes that LG’s OLED panels can produce deeper blacks, but recently set up demonstrations to try and show that those blacks come at the cost of color detail and blown out highlights, the latter of which Sony’s TVs showed no signs of suffering from.

Sony *has* come up with a fresh idea for 2016; it’s a new way of backlighting certain models so that the company can reach truly unreal dimensions. Slim Backlight Drive “uses a unique grid array local dimming and boosting backlighting structure to distribute the backlight source more precisely to each specific zone of the screen.” That’s found on the X930D series, with the flagship 75-inch X940D using full-array backlighting. Just like last year, Google’s Android TV is the underlying foundation for these TVs, so you can count on full support for Google Cast and a wide selection of apps from the Play Store.

Pricing for the various models follows below. Sony’s not playing for the bottom end of the market; that’s clearly Vizio’s territory. Instead, the company says its sets produce the best image quality in the industry. Pretty soon, you’ll be able to sample those claims for yourself at your local Best Buy when they’re next to Samsung, LG, Vizio, and all the rest.

XBR-55X850D, 55″ class (54.6″ diagonal), $2,499.99 MSRP

XBR-65X850D, 65″ class (64.5″ diagonal), $3,499.99 MSRP

XBR-75X850D, 75″ class (74.5″ diagonal), $4,999.99 MSRP

XBR-85X850D, 85″ class (84.6″ diagonal), $9,999.99 MSRP

XBR-55X930D, 55″ class (54.6″ diagonal), $3,299.99 MSRP

XBR-65X930D, 65″ class (64.5″ diagonal), $4,999.99 MSRP

XBR-75X940D, 75″ class (74.5″ diagonal), $7,999.99 MSRP

## 1-Dimensional Array Problem

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/* * * Array is a very simple data structure which is used to store a collection of data, for example roll number of all the students in a class or name of all the countries in the world. To create an array of integer that can hold 10 values, you can write code like this: int[] myList = new int[10]; This problem will test your knowledge on java array. You are given an array of nn integers. A sub-array is "Negative" if sum of all the integers in that sub-array is negative. Count the number of "Negative sub-arrays" in the input array. Note: Subarrays are contiguous chunks of the main array. For example if the array is {1,2,3,5} then some of the subarrays are {1}, {1,2,3}, {2,3,5}, {1,2,3,5} etc. But {1,2,5} is not an subarray as it is not contiguous. Input Format The first line consists an integer nn. The next line will contain nn space separated integers. Value of n will be at most 100. The numbers in the array will range between -10000 to 10000. Output Format Print the answer to the problem. Sample Input 5 1 -2 4 -5 1 Sample Output 9 Explanation These are the ranges of the 9 negative subarrays in this sample: [0:1] [0:3] [0:4] [1:1] [1:3] [1:4] [2:3] [3:3] [3:4] Assume that the index is 0 based. * * */ package liesbeek; import java.util.Arrays; import java.util.Scanner; public class OneDimen { static int count = 0; static int sum = 0; public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ int[] a = new int[5]; Scanner scanner = new Scanner(System.in); int n = scanner.nextInt(); for(int i=0;i<n;i++) { a[i] = scanner.nextInt(); } for(int m=0; m<a.length; m++) { for(n=m+1; n<=a.length; n++) { int[] subArray = Arrays.copyOfRange(a, m, n); calculate(subArray); } } System.out.println(count); } static void calculate(int[] array) { for(int k=0; k<array.length;k++) { sum = sum + array[k]; } if(sum < 0) { ++count; sum = 0; } sum = 0; } } |