题意翻译 给你一个数字,将其“四舍六入”,末尾为5舍去或进位都可,求最终的数字. 题目描述 Vasya has a non-negative integer n n n . He wants to round it to nearest integer, which ends up with 0 0 0 . If n n n already ends up with 0 0 0 , Vasya considers it already rounded. For example, if n=472…

thanks to http://stackoverflow.com/questions/2144459/using-scanf-to-accept-user-input and http://stackoverflow.com/questions/456303/how-to-validate-input-using-scanf for the i/o part. thanks to http://www.haodaima.net/art/137347 for the rounding part…

C. Jeff and Rounding time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to sli…

C. Jeff and Rounding time limit per test:  1 second memory limit per test: 256 megabytes input: standard input output: standard output Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided t…

A. Rounding time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, V…

SAP MM 物料主数据MRP2 视图Rounding Value字段 如下物料号,MRP2视图中,维护了rounding value字段值为50. MRP type :PD Lot size:EX Minmum lot size:500 Rounding value:50 Safety stock:515 MD04结果, MD02 run MRP, 再去看MD04, 触发的PR数量是550, 为什么不是515?因为我们维护了rounding value(最小包装量),所以需求数量是515,即1…

我最开始其实是由于疑惑ROUND_FLOOR和 ROUND_DOWN的表现区别才看了一波文档,但是感觉拉出一票以前没有留意过的东西. 贴一个decimal文档里面的解释: ROUND_CEILING (towards Infinity), ROUND_DOWN (towards zero), ROUND_FLOOR (towards -Infinity), ROUND_HALF_DOWN (to nearest with ties going towards zero), ROUND_HALF_…

http://codeforces.com/problemset/problem/351/A 题意: 2*n个数,选n个数上取整,n个数下取整 最小化 abs(取整之后数的和-原来数的和) 先使所有的数都下取整,累积更改的sum 那么选1个小数上取整,就会使sum-1 整数上下取整不会产生影响 所以有1个整数就可以使上取整的小数少1个 所以ans=min(abs(i-sum)) i∈[n-整数个数,n] #include<cmath> #include<cstdio> #inclu…

  A. Rounding   time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with …

A. Rounding time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, V…

B. Weird Rounding time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k. In the given number of n Polycarp…

BigDecimal 舍入模式(Rounding mode)介绍 什么样的经历,才能领悟成为架构师? >>>   1 RoundingMode介绍 package java.math; public enum RoundingMode {     UP(BigDecimal.ROUND_UP),     DOWN(BigDecimal.ROUND_DOWN),     CEILING(BigDecimal.ROUND_CEILING),     FLOOR(BigDecimal.ROUN…

2.3 Organizing Quantitative Data group quantitative data: To organize quantitative data, we first group the observations into classes (also known as categories or bins) single-value grouping |limit grouping|cutpoint grouping (1)Singlevalue grouping:…

double值由外部传入 private void Compare(double value) { string text; ) //小数位后保留2位 { //小数点后保留2位小数 text = string.Format("{0:0.00}", value); } else { text = "; } } http://stackoverflow.com/questions/5658799/is-it-wrong-to-compare-a-double-to-0-like-…

题目链接:http://codeforces.com/contest/351/problem/A 算法思路:2n个整数,一半向上取整,一半向下.我们设2n个整数的小数部分和为sum. ans = |A - B|; sum = A +(n-b)-B; 所以ans = |sum - (n-b) |; 只有b未知,只需要枚举一下b就得到答案. #include<cstdio> #include<iostream> #include<algorithm> #include<…

http://codeforces.com/contest/352/problem/C 题意:给予N*2个数字,改变其中的N个向上进位,N个向下进位,使最后得到得数与原来数的差的绝对值最小 对每一个浮点数都取其下限,得到的差值就是所有浮点数小数部分的和:然后则需要从2*n个数里面选出n个数取其上限,即下限加1,而如果这个数是个整数,那么不需要加1:因此统计加1个数的上限和下限即可:然后选择min abs(小数部分的和-加1的个数): #include <cstdio> #include <…

这个错误就是精度丢失问题 https://blog.csdn.net/qq496013218/article/details/70792655…

给予N*2个数字,改变其中的N个向上进位,N个向下进位,使最后得到得数与原来数的差的绝对值最小 考虑小数点后面的数字,如果这些数都非零,则就是  abs(原数小数部分相加-1*n), 多一个0 则 min( abs(原数小数部分相加-1*n) ,abs(原数小数部分相加-1*(n-1)) )以此类推 #include<stdio.h> int abs(int a){ )return -a; else return a; } int Min(int a,int b){ if(a<b)ret…

简单题. 删去结尾的不是$0$的数字,保证结尾连续的$k$个都是$0$,如果不能做到,就保留一个$0$. #include<map> #include<set> #include<ctime> #include<cmath> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstring> #in…

暴搜 #include<cstdio> #include<algorithm> using namespace std; int n,K,Div=1,a[21],m,ans=100; bool vis[21]; void calc(int now) { int t=0; bool flag=0; for(int i=m;i>=1;--i) if(!vis[i]) { if((!flag) && a[i]==0) return; t=t*10+a[i]; fla…

Jeff got 2n real numbers a1, a2, ..., a2n as a birthday present. The boy hates non-integer numbers, so he decided to slightly "adjust" the numbers he's got. Namely, Jeff consecutively executes n operations, each of them goes as follows: choose i…

题意: Polycarp is crazy about round numbers. He especially likes the numbers divisible by 10k. In the given number of n Polycarp wants to remove the least number of digits to get a number that is divisible by 10k. For example, if k = 3, in the number 3…

http://codeforces.com/problemset/problem/351/C 题意:有2*n个浮点数a1,a2,a3...a2*n,把他们分成n队,对于每对<A,B>,对A做floor() 操作,对B 做ceil()操作.生成b1...b2*n, 求|(b1+b2+...+b2*n)-(a1+a2+a3...+a2*n)|的最小值. 对于每个数ai,对他们做floor()的cost是up()=ai-floor(ai) ,做ceil()的cost是down()=ceil(ai)-…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 模拟 [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the solution is right? At least,the main ideal 4.use the puts("") or putchar() or printf…

[题目链接]:http://codeforces.com/contest/779/problem/B [题意] 问你要删掉几个数字才能让原来的数字能够被10^k整除; [题解] /* 数字的长度不大; 让你删掉最小的数字个数 使得这个数字能被10^k整除; 搜索; 枚举哪些数字被删掉了; bool记录; 最后再全部乘起来; 枚举删掉1个,删掉两个即可.3.4...; 注意前导0只能有一个的情况就好 */ [完整代码] #include <bits/stdc++.h> using namespa…

1 异常sql upsert into WMBIGDATA.ODS_ES_CHARGING_STATION(id,evcosType,address,serviceTel,supportOrder,operatorId,stationId,stationName,stationTel,stationLng,stationLat,stationStatus,countryCode,stationType,construction,equipmentOwnerId,openAllDay,paymen…

题意 有一个含有\(2n(n \leqslant2000)\)个实数的数列,取出\(n\)个向上取整,另\(n\)个向下取整.问取整后数列的和与原数列的和的差的绝对值. 就是说,令\(a\)为原数列,\(b\)为取整后数列,求 \[ min(abs(\sum_{i=1}^{2n}a-\sum_{i=1}^{2n}b)) \] 解题思路 刚开始大力猜了一波贪心结论,然后怒WA n发-- 我也不知道怎么会想到以下这个奇怪的结论的-- 如果我们设\(n\)个被向上取整的数的小数部分的和为\(a\)(其…

题目出处:http://codeforces.com/problemset/problem/898/A 题目大意:找一个数最近的整十的数 #include<iostream> using namespace std; int main(){ int a,b; cin>>a; b=a; ){ ==){ cout<<a<<endl;; } ==){ cout<<b<<endl;; } a++;b--; } ; } 没啥好考虑的,直接双向查…

在之前的两篇博文分别介绍了常用的hash方法([Data Structure & Algorithm] Hash那点事儿)以及局部敏感hash算法([Algorithm] 局部敏感哈希算法(Locality Sensitive Hashing)),本文介绍的SimHash是一种局部敏感hash,它也是Google公司进行海量网页去重使用的主要算法. 1. SimHash与传统hash函数的区别 传统的Hash算法只负责将原始内容尽量均匀随机地映射为一个签名值,原理上仅相当于伪随机数产生算法.传统…

ORA-00001: 违反唯一约束条件 (.) ORA-00017: 请求会话以设置跟踪事件 ORA-00018: 超出最大会话数 ORA-00019: 超出最大会话许可数 ORA-00020: 超出最大进程数 () ORA-00021: 会话附属于其它某些进程:无法转换会话 ORA-00022: 无效的会话 ID:访问被拒绝 ORA-00023: 会话引用进程私用内存:无法分离会话 ORA-00024: 单一进程模式下不允许从多个进程注册 ORA-00025: 无法分配  ORA-00026:…