{"id":29952,"date":"2023-01-12T07:15:41","date_gmt":"2023-01-12T01:45:41","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-can-someone-examine-the-algorithm-closed\/"},"modified":"2023-01-12T07:15:41","modified_gmt":"2023-01-12T01:45:41","slug":"solved-can-someone-examine-the-algorithm-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-can-someone-examine-the-algorithm-closed\/","title":{"rendered":"[Solved] can someone examine the algorithm [closed]"},"content":{"rendered":"

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You problem is with floating point precision.<\/p>\n

float i = 4.2;\ni *= 100;\nprintf(\"%f\\n\", i);\n<\/code><\/pre>\n
Print: 419.999969<\/code> and not 4.2<\/code> as it should, in this case 419<\/code> is the value used in the coin problem resulting in 22 coins used 16 of 25<\/code>, 1 of 10<\/code>, 1 of 5<\/code> and 4 of 1<\/code> = total 22<\/code><\/p>\n
Use: i = round(i * 100);<\/code> instead of i = i * 100;<\/code><\/p>\n
You need to contemplate the equal value, eg: else if (i >= 10 && i < 25), the range is [ini, fin), close at the beginning, open at the end. If you change the else if<\/code> by if<\/code> you don’t need the while loop<\/code>. <\/p>\n
Final code:<\/p>\n
#include <stdio.h>\n#include <math.h>\n\nint main() {\n    int c = 0;\n    float iv = 4.2;\n    int i = round(iv * 100);\n\n    printf(\"%d\\n\", i);\n    if (i >= 25) {\n        c += i \/ 25;\n        i = i % 25;\n    }\n    if (i >= 10) {\n        c += i \/ 10;\n        i = i % 10;\n    }\n    if (i >= 5) {\n        c += i \/ 5;\n        i = i % 5;\n    }\n    if (i > 0) {\n        c += i;\n    }\n    printf(\"%d\\n\", c);\n\n    return 0;\n}\n<\/code><\/pre>\n
More info about 
\nWhat Every Programmer Should Know About Floating-Point Arithmetic<\/a><\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            <\/span> <\/p><\/div>\n<\/div>\n
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solved can someone examine the algorithm [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] You problem is with floating point precision. float i = 4.2; i *= 100; printf(“%f\\n”, i); Print: 419.999969 and not 4.2 as it should, in this case 419 is the value used in the coin problem resulting in 22 coins used 16 of 25, 1 of 10, 1 of 5 and 4 of 1 … 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":[457,324],"class_list":["post-29952","post","type-post","status-publish","format-standard","hentry","category-solved","tag-algorithm","tag-c"],"yoast_head":"\n[Solved] can someone examine the algorithm [closed] - 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-can-someone-examine-the-algorithm-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] can someone examine the algorithm [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] You problem is with floating point precision. float i = 4.2; i *= 100; printf("%fn", i); Print: 419.999969 and not 4.2 as it should, in this case 419 is the value used in the coin problem resulting in 22 coins used 16 of 25, 1 of 10, 1 of 5 and 4 of 1 ... 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