diff --git a/8.16/N-HarmonicNumber/Harmonic.cpp b/8.16/N-HarmonicNumber/Harmonic.cpp
index 61b2975..b5cb131 100644
--- a/8.16/N-HarmonicNumber/Harmonic.cpp
+++ b/8.16/N-HarmonicNumber/Harmonic.cpp
@@ -46,7 +46,7 @@ int main()
  		{
  			ans+=(1.0/i);
 		}
- 		printf("Case %d: %.10llf\n",++t1,ans);
+ 		printf("Case %d: %.10lf\n",++t1,ans);//WA罪魁祸首 lf->double 不是llf 
  	}
 	return 0;
 }
diff --git a/8.16/N-HarmonicNumber/Harmonic2.cpp b/8.16/N-HarmonicNumber/Harmonic2.cpp
new file mode 100644
index 0000000..8b3e137
--- /dev/null
+++ b/8.16/N-HarmonicNumber/Harmonic2.cpp
@@ -0,0 +1,40 @@
+//调和级数近似公式：f(n)=ln(n)+r+1.0/(2*n)
+//r是欧拉常数
+//注意：此公式只适用于n很大的情况。
+#include<cstdio>
+#include<algorithm>
+#include <cmath>
+using namespace std;
+#define r 0.57721566490153286060651209//欧拉常数
+const int N=1e8;
+const int w=1e4;
+double a[w+5];
+int t,n;
+int t1;
+double ans;
+inline void cun()
+{
+	for(int i=1;i<=w;i++)
+	{
+		a[i]=a[i-1]+(1.0/i);
+	} 
+}
+int main()
+{
+ 	cun();
+ 	scanf("%d",&t);
+ 	while(t--)
+ 	{
+ 		scanf("%d",&n);
+ 		if(n<=w)
+ 		{
+ 			printf("Case %d: %.10lf\n",++t1,a[n]);//lf->double 不是llf 
+		}
+ 		else
+ 		{
+ 			ans=log(n)+r+1.0/(2*n);//cmath 中log就是ln 
+			printf("Case %d: %.10lf\n",++t1,ans);
+		}  		
+ 	}
+	return 0;
+}
diff --git a/8.16/N-HarmonicNumber/Harmonicreference.cpp b/8.16/N-HarmonicNumber/Harmonicreference.cpp
new file mode 100644
index 0000000..9adac22
--- /dev/null
+++ b/8.16/N-HarmonicNumber/Harmonicreference.cpp
@@ -0,0 +1,34 @@
+#include<cstdio>
+#include<cstring>
+#include<algorithm>
+using namespace std;
+ 
+double a[2500050];
+ 
+void init()
+{
+    double sum=0.0;
+    for(int i=1; i<100000001; i++)
+    {
+        sum+=1.0/i;
+        if(i%50==0) a[i/50]=sum;  //每隔49个记录一次。
+    }
+}
+ 
+int main()
+{
+    init();
+    int t,flag=1;
+    scanf("%d",&t);
+    while(t--)
+    {
+        int n;
+        scanf("%d",&n);
+        int f=n/50;
+        double ans=a[f];
+        for(int i=50*f+1; i<=n; i++)  //很明显循环在50以内
+            ans+=1.0/i;
+        printf("Case %d: %.10lf\n",flag++,ans);
+    }
+    return 0;
+}
diff --git a/8.16/N-HarmonicNumber/Harmonicreference1-2.cpp b/8.16/N-HarmonicNumber/Harmonicreference1-2.cpp
new file mode 100644
index 0000000..dd66d53
--- /dev/null
+++ b/8.16/N-HarmonicNumber/Harmonicreference1-2.cpp
@@ -0,0 +1,42 @@
+#include<iostream>
+#include<cstdio>
+using namespace std;
+
+const int N=1e6+5;
+const int size=100;
+
+double hn[N];
+
+void gethn(){
+	int i,j,k;
+	double temp;
+	hn[0]=0;
+	temp=0;
+	k=1;
+	for(i=1;i<=N;i++){
+		for(j=1;j<=size;j++){
+		temp+=(1.0)/k;	
+		k++;
+		}
+		hn[i]=temp;
+	}
+}
+
+double readhn(int x){
+	double ans;
+	int c=x%size;
+	ans=hn[x/size];
+	for(int i=x-c+1;i<=x;i++) ans+=1.0/i;
+	return ans;
+}
+
+int main(){
+	gethn();
+	int i,t,n; 
+	scanf("%d",&t);
+	for(i=1;i<=t;i++){
+		scanf("%d",&n);
+		printf("Case %d: %.10lf\n",i,readhn(n));
+	}
+	return 0;
+}
diff --git a/8.16/N-HarmonicNumber/Harmonicreference2.cpp b/8.16/N-HarmonicNumber/Harmonicreference2.cpp
new file mode 100644
index 0000000..97edb6d
--- /dev/null
+++ b/8.16/N-HarmonicNumber/Harmonicreference2.cpp
@@ -0,0 +1,34 @@
+#include<stdio.h>
+#include<string.h>
+#include<algorithm>
+#include<math.h>
+#define r 0.57721566490153286060651209//欧拉常数
+using namespace std;
+double a[10010];
+int main()
+{
+    a[1]=1.0;
+    for(int i=2; i<10000; i++)
+        a[i]=a[i-1]+1.0/i;
+    int t;
+    scanf("%d",&t);
+    int o=1;
+    while(t--)
+    {
+        int n;
+        double res=0.0;
+        scanf("%d",&n);
+        if(n<10000)
+        {
+            printf("Case %d: ",o++);
+            printf("%.10lf\n",a[n]);
+        }
+        else
+        {
+            printf("Case %d: ",o++);
+            double t=log(n)+r+1.0/(2*n);
+            printf("%.10lf\n",t);
+        }
+    }
+    return 0;
+}
diff --git a/8.16/N-HarmonicNumber/Readme.md b/8.16/N-HarmonicNumber/Readme.md
index cc02b16..baa30f7 100644
--- a/8.16/N-HarmonicNumber/Readme.md
+++ b/8.16/N-HarmonicNumber/Readme.md
@@ -1,12 +1,20 @@
-# 题目
-* 链接
+# N - Harmonic Number
+* https://vjudge.net/contest/509210#problem/N
 ### 题意
-
-### 做法
-
+h(n)=1/1+1/2+……+1/n;n<=1e8
+### 做法(2种)
+对应两个代码
+1.分层打表：每50/100个记录一次，查的时候再接着算
+2.调和级数近似公式：前1e4打表，后面的调和级数近似公式（因为此公式只适用于n很大的情况）
 ### 关键词
-
+基础数论、调和级数近似公式、分层打表、欧拉常数
 ### 易错点
-* 
+* //lf->double 不是llf 
+* //cmath 中log就是ln 
+* 直接用int/long long扩大倍数打表
 ### 工具箱
-* 
\ No newline at end of file
+* doubl&%lf https://blog.csdn.net/weixin_46585199/article/details/119044763
+* 调和级数近似公式：f(n)=ln(n)+r+1.0/(2*n),r是欧拉常数
+//注意：此公式只适用于n很大的情况。
+* #define r 0.57721566490153286060651209//欧拉常数
+