8.18下午

8.16/I-GCD/doc/Readme.md

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+# 

8.16/I-GCD/main.cpp

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+#include <iostream>
+#include<cstdio>
+#include<algorithm>
+#include <cmath>
+using namespace std;
+int main()
+{
+    int t;
+    scanf("%d",&t);
+    while(t--)
+    {
+        int n;
+        scanf("%d",&n);
+        for(int i=1;i<=n; i++)
+        {
+
+        }
+    }
+    return 0;
+}

8.16/I-GCD/test/out.txt

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+## z:\Chao\src\Template\test\in.txt 
+2020/03/14 ÖÜÁù 11:41:28.68 
+Hello Easy C++ project!
+
+-----------------------------------------------
+Process exited after 200 ms with return value 0
+

8.16/J-LeadingandTrailing/Readme.md

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+# J - Leading and Trailing
+* https://vjudge.net/contest/509210#problem/J
+### 题意
+n^k的前3位、后3位n (2 ≤ n < 2^31) and k (1 ≤ k ≤ 10^7).
+### 做法
+1. 后三位快速幂
+2. 前三位10^ (log10(n^k)==k*log10(n)),整数部分决定0的个数，小数部分决定数字 *100取前三位
+### 关键词
+快速幂、求很大的n^k、log10、pow、modf
+### 易错点
+* 快速幂中now=n%mod;//要mod否则会太大10e6*10e6
+* 快速幂忘return
+* //%03d 001 000 补齐1前面的0
+* //没有（int)导致没有输出
+### 工具箱
+* log10(n^k)==k*log10(n)
+* pow(10,cur)==10^cur
+* %03d 001 000 整数用零补齐
+* y=modf(n,&x);n的小数部分给y,整数部分给x
+* 题解 https://www.cnblogs.com/KirinSB/p/9409120.html

8.16/J-LeadingandTrailing/doc/Readme.md

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+# 

8.16/J-LeadingandTrailing/test/out.txt

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+## z:\Chao\src\Template\test\in.txt 
+2020/03/14 ÖÜÁù 11:41:28.68 
+Hello Easy C++ project!
+
+-----------------------------------------------
+Process exited after 200 ms with return value 0
+