Japt, (削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
###Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
##8-byte Solution: õ å+ è\U
###Explanation: õ å+ è\U õ // Create a range from [1...Input] å+ // Cumulative reduce by addition è // Return the number of occurrences where: \U // Item == Input
##10-byte Solution:
õ å+ aU >JøU
###Explanation: õ å+ aU >JøU õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aUøU // Return the last index ofDoes it contain the input >J // > -1?
Japt, (削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
###Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
##8-byte Solution: õ å+ è\U
###Explanation: õ å+ è\U õ // Create a range from [1...Input] å+ // Cumulative reduce by addition è // Return the number of occurrences where: \U // Item == Input
##10-byte Solution:
õ å+ aU >J
###Explanation: õ å+ aU >J õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aU // Return the last index of the input >J // > -1
Japt, (削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
###Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
õ å+ øU
###Explanation: õ å+ øU õ // Create a range from [1...Input] å+ // Cumulative reduce by addition øU // Does it contain the input?
Japt, (削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
##Explanation###Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
###10##8-byte Solution: õ å+ è\U
###Explanation: õ å+ è\U õ // Create a range from [1...Input] å+ // Cumulative reduce by addition è // Return the number of occurrences where: \U // Item == Input
##10-byte Solution:
õ å+ aU >J
##Explanation###Explanation: õ å+ aU >J õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aU // Return the last index of the input >J // > -1
Japt, (削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
##Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
###10-byte Solution:
õ å+ aU >J
##Explanation: õ å+ aU >J õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aU // Return the last index of the input >J // > -1
Japt, (削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
###Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
##8-byte Solution: õ å+ è\U
###Explanation: õ å+ è\U õ // Create a range from [1...Input] å+ // Cumulative reduce by addition è // Return the number of occurrences where: \U // Item == Input
##10-byte Solution:
õ å+ aU >J
###Explanation: õ å+ aU >J õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aU // Return the last index of the input >J // > -1
Japt, 10(削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
##Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
###10-byte Solution:
õ å+ aU >J
##Explanation: õ å+ aU >J õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aU // Return the last index of the input >J // > -1
After we cumulatively reduce the range [1...Input] by addition, we check to see if the Input exists in that range.
Japt, 10 bytes
õ å+ aU >J
##Explanation: õ å+ aU >J õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aU // Return the last index of the input >J // > -1
After we cumulatively reduce the range [1...Input] by addition, we check to see if the Input exists in that range.
Japt, (削除) 10 (削除ここまで) 7 bytes
Saved 3 bytes thanks to @Luke and @ETHproductions
*8Ä ¬v1
##Explanation: *8Ä ¬v1 ¬ // Square root of: *8 // Input * 8 Ä // +1 v1 // Return 1 if divisible by 1; Else, return 0
###10-byte Solution:
õ å+ aU >J
##Explanation: õ å+ aU >J õ // Create a range from [1...Input] å+ // Cumulative reduce by addition aU // Return the last index of the input >J // > -1