Fun with Integers -codeforces
讀懂題意就好,有個最直接的想法,列舉\(a,b\),複雜度\(O(n^2)\)。然而換成列舉\(a,x\),複雜度\(O(nlg(n))\)。\(<a,b>\)和\(<b,a>\)的貢獻只能算一次。
程式碼
#include <bits/stdc++.h> using namespace std; int n; long long ans; int main() { cin >> n; for (int a = 2; a <= n / 2; ++a) for (int x = 2; x * a <= n; ++x) ans += x * 4; cout << ans << endl; return 0; }
相關推薦
Fun with Integers -codeforces
讀懂題意就好,有個最直接的想法,列舉\(a,b\),複雜度\(O(n^2)\)。然而換成列舉\(a,x\),複雜度\(O(nlg(n))\)。\(<a,b>\)和\(<b,a>\)的貢獻只能算一次。 程式碼 #include <bits/stdc++.h> usin
Fun with Integers CodeForces - 1062D
http://codeforces.com/contest/1062/problem/D 考慮每個乘數x的貢獻 x=2時 (+-2,+-4) (+-3,+-6) (+-4,+-8)... x=3時 (+-2,+-6) (+-3,+-9) (+-4,+-12)...
Codeforces Round #520 (Div. 2) D. Fun with Integers
long with 復雜度 include 就是 cstring ger test pan D. Fun with Integers 題目鏈接:https://codeforc.es/contest/1062/problem/D 題意: 給定一個n,對於任意2<=|a
D. Fun with Integers
保存 sync http uri pro ans sin mes spa 鏈接 [http://codeforces.com/contest/1062/problem/D] 題意 給你n,讓你從2到n這個區間找任意兩個數,使得一個數是另一個的因子,絕對值小的可以變為絕對值大
CF1062D Fun with Integers
思路: 找規律。 實現: 1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int main() 5 { 6 ll n; 7 while (cin >
Fun with Integers
You are given a positive integer nn greater or equal to 22 . For every pair of integers aa and bb (2≤|a|,|b|≤n2≤|a|,|b|≤n ), you can transform aa into
poj3511--A Simple Problem with Integers(線段樹求和)
poj pac style som can com onos roman miss A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K
線段樹專題 POJ3468 A Simple Problem with Integers
strong print style update else algo linker clas uil 題意:n個點。m個操作。兩種操作類型。C X Y K 表示區間[x,y]上每一個點值加k。Q X Y 求區間[x,y]的和 分析:線段樹區間求和,裸模板 註意
poj 3468 A Simple Problem with Integers
arch fin uil range swe next char () limit A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total
poj 3468 A Simple Problem with Integers(原來是一道簡單的線段樹區間修改用來練練splay)
long 兩個 可能 style push ios stream 區間 pan 題目鏈接:http://poj.org/problem?id=3468 題解:splay功能比線段樹強大當然代價就是有些操作比線段樹慢,這題用splay實現的比線段樹慢上一倍。線段樹用l
POJ3468_A Simple Problem with Integers(線段樹/成段更新)
div sub accepted scan can print onos cst align 解題報告 題意: 略 思路: 線段樹成段更新,區間求和。 #include <iostream> #include <cstring> #
A Simple Problem with Integers POJ - 3468
nts sub other numbers opera clu ios mes initial You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One
POJ-3468 A Simple Problem with Integers(線段樹、段變化+段查詢、模板)
sum .org miss numbers ... bsp wid scanf accepted A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Su
POJ - 3468 A Simple Problem with Integers
pan const value int stream put code all pda You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One typ
POJ 3468 A Simple Problem with Integers 線段樹成段更新
sca query string int open print ring cnblogs pac 線段樹功能 update:成段更新 query:區間求和 #include <iostream> #include <string>
hdu 4267 A Simple Problem with Integers
pre pri ets using type main return ger 如果 題意:給出一個長為n的序列,然後有q個操作,有兩種操作方式 1.輸入a,b,k,c,表示將區間[a,b]中的數i滿足(i-a)%k == 0加上c. 2.輸入一個數y,輸出序列中第
POJ 3468 A Simple Problem with Integers(線段樹 單點更新+區間求和 )
names || log shu 更新 can pro struct sim 題目鏈接:http://poj.org/problem?id=3468 題意:單點更新,區間求和。 題解:裸題 1 //POJ 3468 A Simple Problem with
1023. Have Fun with Numbers
for each stand test case pro 時間 cas 內存 app ring 1023. Have Fun with Numbers (20) 時間限制 400 ms 內存限制 65536 kB 代碼長度限制 16000 B 判題程序 St
poj3468 A Simple Problem with Integers
strong sin getch har post query tree col turn 思路: 線段樹區間更新。 實現: 1 #include <cstdio> 2 #include <cstring> 3 using namespace
A Simple Problem with Integers~POJ - 3468
font nts eal int mil 模板 some ber blog You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of