微小的更新

F0RE1GNERS · Oct 26, 2018 · ae864e0 · ae864e0
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diff --git a/2-数学.md b/2-数学.md
@@ -1249,12 +1249,10 @@ LL exBSGS(LL a, LL b, LL p) { // a^x = b (mod p)
 
 ## 数论分块
 
-令 $g(i) = \lfloor \frac{n}{i} \rfloor$。在求 $g(i)$ 的同时求出 $[i,j]$ 的 $g(i)$。那么 $j=n/(n/i)$。
+$f(i) = \lfloor \frac{n}{i} \rfloor=v$ 时 $i$ 的取值范围是 $[l,r]$。
 
 ```cpp
-for (LL i = 1; i <= n; i = j + 1) {
-    j = n / (n / i);
-    ans += (j - i + 1) * (n / i);
+for (LL l = 1, v, r; l <= N; l = r + 1) {
+    v = N / l; r = N / v;
 }
 ```
-
diff --git a/template.pdf b/template.pdf