diff --git a/TangDou/Topic/Mobius/P4318.cpp b/TangDou/Topic/Mobius/P4318.cpp
index 61ddbfc..753b170 100644
--- a/TangDou/Topic/Mobius/P4318.cpp
+++ b/TangDou/Topic/Mobius/P4318.cpp
@@ -5,7 +5,7 @@ using namespace std;
 const int N = 100010;
 
 // 筛法求莫比乌斯函数(枚举约数)
-int mu[N], sum[N]; // sum[N]:梅滕斯函数,也就是莫比乌斯函数的前缀和
+int mu[N];
 int primes[N], cnt;
 bool st[N];
 void get_mobius(int n) {
@@ -25,14 +25,11 @@ void get_mobius(int n) {
             mu[t] = -mu[i];
         }
     }
-    // 维护u(x)前缀和:梅滕斯函数
-    for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + mu[i];
 }
-int check(int g, int k) {
+int check(int mid, int k) {
     int res = 0;
-    for (int i = 1; i * i <= g; i++)
-        res = res + mu[i] * (g / i / i);
-
+    for (int i = 1; i * i <= mid; i++) // 枚举范围内每个平方数
+        res += mu[i] * (mid / i / i);
     return res >= k;
 }
 
@@ -45,13 +42,13 @@ signed main() {
     while (T--) {
         cin >> k; // 第k个数
         int l = 1, r = 2e9;
-        while (l <= r) {
+        while (l < r) {
             int mid = l + r >> 1;
-            if (!check(mid, k))
-                l = mid + 1;
+            if (check(mid, k))
+                r = mid;
             else
-                r = mid - 1;
+                l = mid + 1;
         }
-        cout << r + 1 << endl;
+        cout << r << endl;
     }
 }
diff --git a/TangDou/Topic/莫比乌斯函数.md b/TangDou/Topic/莫比乌斯函数.md
index 43ece3c..4cb1bd4 100644
--- a/TangDou/Topic/莫比乌斯函数.md
+++ b/TangDou/Topic/莫比乌斯函数.md
@@ -208,64 +208,59 @@ signed main() {
 > **注**: 与上一题的区别在于上一题明确给出了最大值$n$,也就是右边界的范围，本题没有告诉我们范围，需要我们自己找到右边界。随着右边界越来越大，肯定符合条件的数字个数也会越来越多，也就是上面说到的单调性。我们可以用二分来假设一个右边界，然后不断的收缩区间来找到准备的右边界:在上道题的基础上加上二分，判断$1$到$mid$是否有$K$个无平方因子的数，以此改变左右边界即可。
 
 ```cpp {.line-numbers}
-#include<iostream>
-#define ll long long 
+#include <bits/stdc++.h>
 using namespace std;
-int vis[40560],mo[40560],p[4253],n;
-void init()
-{
-    int tot=0,k;
-    mo[1]=1;
-    for(int i=2;i<=40559;i++)
-    {
-        if(vis[i]==0)
-        {
-            p[++tot]=i;
-            mo[i]=-1;
+#define int long long
+#define endl "\n"
+const int N = 100010;
+
+// 筛法求莫比乌斯函数(枚举约数)
+int mu[N];
+int primes[N], cnt;
+bool st[N];
+void get_mobius(int n) {
+    mu[1] = 1;
+    for (int i = 2; i <= n; i++) {
+        if (!st[i]) {
+            primes[cnt++] = i;
+            mu[i] = -1;
         }
-        for(int j=1;j<=tot&&(k=i*p[j])<=40559;j++)
-        {
-            vis[k]=1;
-            if(i%p[j]!=0)mo[k]=-mo[i];
-            else 
-            {
-                mo[k]=0;
+        for (int j = 0; primes[j] <= n / i; j++) {
+            int t = primes[j] * i;
+            st[t] = true;
+            if (i % primes[j] == 0) {
+                mu[t] = 0;
                 break;
             }
+            mu[t] = -mu[i];
         }
     }
 }
-bool judge(int x)
-{
-	ll ans=0;int i;
-    for(int i=1;i*i<=x;i++)
-    {
-        ans+=mo[i]*(x/(i*i));
-    }
-    if(ans>=n)return true;
-    else return false;
+int check(int mid, int k) {
+    int res = 0;
+    for (int i = 1; i * i <= mid; i++) // 枚举范围内每个平方数
+        res += mu[i] * (mid / i / i);
+    return res >= k;
 }
-int main()
-{
-    int t;
-    ll l,r,mid;
-    init();
-    scanf("%d",&t);
-    while(t--)
-    {
-        scanf("%d",&n);
-        l=n,r=1644934082;
-        while(l<r) 
-        {
-            // cout<<l<<" oo "<<r<<endl;
-			mid=(l+r)>>1;
-			if(judge(mid)) 
-                r=mid;
-			else l=mid+1;
-		}
-        printf("%lld\n",r);
+
+signed main() {
+    // 获取莫比乌斯函数值
+    get_mobius(N - 1);
+
+    int T, k;
+    cin >> T; // T组测试数据
+    while (T--) {
+        cin >> k; // 第k个数
+        int l = 1, r = 2e9;
+        while (l < r) {
+            int mid = l + r >> 1;
+            if (check(mid, k))
+                r = mid;
+            else
+                l = mid + 1;
+        }
+        cout << r << endl;
     }
-    return 0;
 }
 ```