整除分块

模板：P1403 [AHOI2005]约数研究 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

 1 #include "bits/stdc++.h"
 2 #define PII pair<int,int>
 3 #define rep(i,z,n) for(int i = z;i <= n; i++)
 4 #define per(i,n,z) for(int i = n;i >= z; i--)
 5 #define ll long long
 6 #define db double
 7 #define vi vector<int>
 8 #define debug(x) cerr << "!!!" << x << endl;
 9 using namespace std;
10 //从某个串中把某个子串替换成另一个子串
11 string& replace_all(string& src, const string& old_value, const string& new_value) {
12     // 每次重新定位起始位置，防止上轮替换后的字符串形成新的old_value
13     for (string::size_type pos(0); pos != string::npos; pos += new_value.length()) {
14         if ((pos = src.find(old_value, pos)) != string::npos) {
15             src.replace(pos, old_value.length(), new_value);
16         }
17         else break;
18     }
19     return src;
20 }
21 inline ll read()
22 {
23     ll s,r;
24     r = 1;
25     s = 0;
26     char ch = getchar();
27     while(ch < '0' || ch > '9'){
28         if(ch == '-')
29             r = -1;
30         ch = getchar();
31     }
32     while(ch >= '0' && ch <= '9'){
33         s = (s << 1) + (s << 3) + (ch ^ 48);
34         ch = getchar();
35     }
36     return s * r;
37 }
38 inline void write(ll x)
39 {
40     if(x < 0) putchar('-'),x = -x;
41     if(x > 9) write(x / 10);
42     putchar(x % 10 + '0');
43 }
44 int main()
45 {
46     int n;
47     n = read();
48     int nex = 1;
49     int bnd;
50     int ans = 0;
51     int res;
52     rep(i,1,n){
53         res = n / nex;
54         if(res)//防止除0
55             bnd = n / res;
56         ans += res * (bnd - nex + 1);//为res的除数一共有bnd - nex + 1个
57         nex = bnd + 1;
58     }
59     cout << ans;
60     return 0;
61 }

 

posted @ 2022-03-15 16:26  scannerkk  阅读(27)  评论(0)    收藏  举报