覺醒技能-屬性強化測試報告 - 手遊

By James
at 2013-11-04T15:27
at 2013-11-04T15:27
Table of Contents
剛剛在跟朋友測試 計算屬性強化覺醒技的效益
翻了舊文 借個圖來改
在這記錄一下 也給大家參考
以下用比較實際的12顆同色珠來計算
A - 4串
xxxxxx
xxxooo
oooxxx
xxxooo
oooxxx
公式是 n * (1 + 0.25*(c-1)), n為強化色串數, c為combo數
因此4串4c的倍數會是
4 * (1 + 0.25*(4-1)) = 7
---
B - 2列
xxxxxx
xxxxxx
oooooo
xxxxxx
oooooo
公式是 1.75 * n * (1 + 0.25*(c-1)) * (1 + 0.1*k*n), n與c同上, k為覺醒數
不同覺醒數下2列2c的倍數會是
2列2c 2覺醒: 1.75 * 2 * (1 + 0.25*(2-1)) * (1 + 0.1*2*2) = 6.125
2列2c 3覺醒: 1.75 * 2 * (1 + 0.25*(2-1)) * (1 + 0.1*3*2) = 7
2列2c 4覺醒: 1.75 * 2 * (1 + 0.25*(2-1)) * (1 + 0.1*4*2) = 7.875
只要3覺醒 倍數就跟A圖一樣了
再考慮額外c的難度 效益應該更大
撒旦一隻就3個真是太過分了啊~_~
--
翻了舊文 借個圖來改
在這記錄一下 也給大家參考
以下用比較實際的12顆同色珠來計算
A - 4串
xxxxxx
xxxooo
oooxxx
xxxooo
oooxxx
公式是 n * (1 + 0.25*(c-1)), n為強化色串數, c為combo數
因此4串4c的倍數會是
4 * (1 + 0.25*(4-1)) = 7
---
B - 2列
xxxxxx
xxxxxx
oooooo
xxxxxx
oooooo
公式是 1.75 * n * (1 + 0.25*(c-1)) * (1 + 0.1*k*n), n與c同上, k為覺醒數
不同覺醒數下2列2c的倍數會是
2列2c 2覺醒: 1.75 * 2 * (1 + 0.25*(2-1)) * (1 + 0.1*2*2) = 6.125
2列2c 3覺醒: 1.75 * 2 * (1 + 0.25*(2-1)) * (1 + 0.1*3*2) = 7
2列2c 4覺醒: 1.75 * 2 * (1 + 0.25*(2-1)) * (1 + 0.1*4*2) = 7.875
只要3覺醒 倍數就跟A圖一樣了
再考慮額外c的難度 效益應該更大
撒旦一隻就3個真是太過分了啊~_~
--
Tags:
手遊
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