{"id":24407,"date":"2022-12-02T19:02:47","date_gmt":"2022-12-02T13:32:47","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-xor-logic-in-python\/"},"modified":"2022-12-02T19:02:47","modified_gmt":"2022-12-02T13:32:47","slug":"solved-xor-logic-in-python","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-xor-logic-in-python\/","title":{"rendered":"[Solved] Xor logic in python"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-40669201\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"40669201\" data-parentid=\"40634603\" data-score=\"1\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p>The trick is to recognize that you don&#8217;t have to test all the <code>a<\/code> up to <code>x<\/code>.  For <code>a^x == a+x<\/code>, then <code>a&amp;x == 0<\/code>.  So we count the number of zeroes in the bitstring of <code>x<\/code> and then out answer is <code>2**count -1<\/code><\/p>\n<pre><code>test = int(input())\nfor _ in range(test):\n    x = int(input())\n    print(2**bin(x)[2:].count('0') -1)\n<\/code><\/pre>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">0<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved Xor logic in python <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] The trick is to recognize that you don&#8217;t have to test all the a up to x. For a^x == a+x, then a&amp;x == 0. So we count the number of zeroes in the bitstring of x and then out answer is 2**count -1 test = int(input()) for _ in range(test): x = int(input()) &#8230; <a title=\"[Solved] Xor logic in python\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-xor-logic-in-python\/\" aria-label=\"More on [Solved] Xor logic in python\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[5174,325,349],"class_list":["post-24407","post","type-post","status-publish","format-standard","hentry","category-solved","tag-bitwise-xor","tag-performance","tag-python"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>[Solved] Xor logic in python - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-xor-logic-in-python\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Xor logic in python - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] The trick is to recognize that you don&#8217;t have to test all the a up to x. For a^x == a+x, then a&amp;x == 0. So we count the number of zeroes in the bitstring of x and then out answer is 2**count -1 test = int(input()) for _ in range(test): x = int(input()) ... 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