From ee5d06b78c87e01ddf0583983175a7969b2bc5b5 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Thu, 14 Dec 2023 11:14:41 +0800
Subject: [PATCH] 'commit'

---
 TangDou/Topic/SpareTire.cpp | 9 ++++++---
 1 file changed, 6 insertions(+), 3 deletions(-)

diff --git a/TangDou/Topic/SpareTire.cpp b/TangDou/Topic/SpareTire.cpp
index 59ba4eb..5899380 100644
--- a/TangDou/Topic/SpareTire.cpp
+++ b/TangDou/Topic/SpareTire.cpp
@@ -7,10 +7,13 @@ const int Six = 166666668; /// 6，2关于mod的乘法逆元
 const int Two = 500000004;
 const int mod = 1e9 + 7; /// 尽量这样定义mod ，减少非必要的麻烦
 
-inline int Mod(int a, int b) {
+int Mod(int a, int b) {
     return (a % mod) * (b % mod) % mod;
 }
-inline int F(int k, int n) { /// 求(k*n)^2+k*n
+
+// 递推函数可以用通项式直接求解
+// 通项式：a[kn]=(k*n)^2+k*n
+int F(int k, int n) { 
     return (Mod(k, k) * Mod(Mod(n, n + 1), Mod(n + n + 1, Six)) % mod + Mod(Mod(1 + n, n), Mod(k, Two))) % mod;
 }
 
@@ -23,7 +26,7 @@ signed main() {
     int n, m;
 
     while (cin >> n >> m) {
-        int sum = F(1, n), ans = 0; /// 计算总和sum
+        int sum = F(1, n), ans = 0; // 计算总和sum
 
         int t = m; // 复制出来
         for (int i = 2; i * i <= t; i++) {