From e3de53982611f716daf67a5677eb031ef2fa7b2c Mon Sep 17 00:00:00 2001
From: shIne_wOng <1551885@tongji.edu.cn>
Date: Fri, 3 May 2019 11:03:45 +0800
Subject: [PATCH] update the display of latex formula

---
 thu_dsa/chp1.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/thu_dsa/chp1.md b/thu_dsa/chp1.md
index 6587bd5..3da2f24 100644
--- a/thu_dsa/chp1.md
+++ b/thu_dsa/chp1.md
@@ -48,7 +48,7 @@ _To measure is to know. If you cannot measure it, you cannot improve it
 
 考虑我们要评价两个不同算法的优劣，一个很自然的思想是实验测量，通过测量这两个算法在同一问题实例(instance)P的情况下，两个算法的运行代价。即
 
-$$T_A(P) = the cost of algorithm A to solve instance P$$
+$$T_A(P) = the_cost_of_algorithm_A_to_solve_instance_P$$
 
 可是，这样的测量真的有意义吗？
 
@@ -56,7 +56,7 @@ $$T_A(P) = the cost of algorithm A to solve instance P$$
 
 那好，我们对不同的问题实例进行抽象。根据经验，一般的算法的运行代价和问题的规模是相关的。一般说来，问题的规模越大，算法运行的代价越大(当然也有反例，如Hailstone)。这样，我们就可以把基于问题实例P的测量，转化为基于一系列规模为n的问题实例P的测量，即
 
-$$ T_A(n) = the cost of algorithm to solve instances of scale n $$
+$$ T_A(n) = the_cost_of_algorithm_to_solve_instances_of_scale_n $$
 
 有下面两种方案：
 
@@ -97,5 +97,5 @@ $$ T_A(n) = the cost of algorithm to solve instances of scale n $$
 
 由于每步基本操作的运行时间是固定的，因此我们可以使用算法执行所需要的基本操作的次数来评价一个算法。这样
 
-$$T(n) = number of basic operations to solve a problem of scale n$$
+$$T(n) = number_of_basic_operations_to_solve_a_problem_of_scale_n$$