CSU 1040: Round-number
Description
Most of the time when rounding a given number, it is customary to round to some multiple of a power of 10. However, there is no reason why we cannot use another multiple to do our rounding to. For example, you could round to the nearest multiple of 7, or the nearest multiple of 3.
Given an int n and an int b, round n to the nearest value which is a multiple of b. If n is exactly halfway between two multiples of b, return the larger value.
相關推薦
CSU 1040: Round-number
Description Most of the time when rounding a given number, it is customary to round to some multiple of a power of 10. However, there is no reason why w
1040 Round-number 數學題
Description Most of the time when rounding a given number, it is customary to round to some multiple of a power of 10. However, th
POJ-3252 Round Number
題意:給你2個數a,b,求[a,b]區間中的數滿足2進位制0>=1的個數. (1≤a<b≤ 2,000,000,000). 思路:用數位DP來解決,dp[i][count0][count1],表示搜尋到第i位的時候,0與1的數量。記憶化搜尋到底的時候如果0的
Python標準庫:內建函式round(number[, ndigits])
本函式是實現對浮點數進行四捨五入的計算。引數number是浮點數;引數ndigits是保留幾位小數,預設是0值。不過要注意的是轉換出來的小數表示可能還是差異,最好使用庫專用的數字來計算四捨五入運算。例子:#round() print('round(0.5)=', round
JD 題目1040:Prime Number (篩法求素數)
rime 簡單 set end std tdi href num mod OJ題目:click here~~ 題目分析:輸出第k個素數 貼這麽簡單的題目,目的不清純 用篩法求素數的基本思想是:把從1開始的、某一範圍內的正整數從小到大順序排列
CodeFroce Round 340 div2 E XOR and Favorite Number【莫隊算法】
logs pairs num ble ber col nod 操作 app 題面: Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query
Codeforces Round #480 (Div. 2) E. The Number Games
hid boa sse next contest sample art oid ssis E. The Number Games time limit per test 3 seconds memory limit per test 256 megabytes in
Codeforces Round #340 (Div. 2) E. XOR and Favorite Number 【莫隊算法 + 異或和前綴和的巧妙】
lag ans integer ons lap stand 推出 open closed 任意門:http://codeforces.com/problemset/problem/617/E E. XOR and Favorite Number time limit p
[csa round#1]Number Elimination——動態規劃+計數 dalaos' blogs Some Links
題目大意: 你有 n n n個方塊排成一排,每個方塊有一個權值
Codeforces Round #491 (Div. 2) E Bus Number —— 排列組合
題目連結:http://codeforces.com/contest/991/problem/E 有重複問題的全排列是AnnAnum[0]num[0]Anum[1]num[1]...Anum[9]num[9]AnnAnum[0]num[0]Anum[1]num
[csa round#1]Number Elimination——動態規劃+計數
題目大意: 你有nnn個方塊排成一排,每個方塊有一個權值aia_iai,你每次可以選擇一個二元組(x,y)x<y(x,y) x<y(x,y)x<y,並消除x和y中權值
【Codeforces Round 340 (Div 2)E】【莫隊演算法 真實區間思想】XOR and Favorite Number m組區間詢問 問區間中多少連續段異或值為k
E. XOR and Favorite Number time limit per test 4 seconds memory limit per test 256 megabytes input standard input output
Codeforces Round #228 (Div. 2), problem: (A) Fox and Number Game
Note In the first example the optimal way is to do the assignment: x2 = x2 - x1. In the second example the optimal sequence of operations is: x3 = x3 -
Codeforces Round #266 (Div. 2) C. Number of Ways
C. Number of Ways time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output
hdu1212 Big Number &第六屆山東省賽Single Round Math (同餘定理,大數取模)
題目大意:每次輸入兩個數,第一個是高精度,第二個數小於100000;求 a mod b 根據同餘定理: (a+b)% c = (a%c+ b%c)%c (a*b)%c = ( a%c* b%c)%
CSU 1427: Infected Computer 1433: Defend the Bases 1436: Revenge of the Round Table 1438: Swyper Key
#include <iostream> #include <algorithm> #include <stdio.h> #define max 20005 #define ll long long using namespace std;
BestCoder Round #91
output focus block contain scrip tin int rip cte A.Lotus and Characters(技巧 + 思維) Problem Description Lotus has n kinds of characters,ea
消失的數字(number)
哪些 code b+ 描述 橡皮擦 什麽 正序 for 消失 消失的數字(number) Time Limit:1000ms Memory Limit:128MB 題目描述 rsy擁有n個數,這n個數分別是a1,a2,…,an。 後來出現了一個熊孩子zhw,用橡皮擦
Oracle數據庫中number類型在java中的使用
integer big ora col acl number bsp java tex 1)如果不指定number的長度,或指定長度n>18 id number not null,轉換為pojo類時,為java.math.BigDecimal類型 2)如果num