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inquiry
02:Integer
Inquiry
#include #include usingnamespacestd; intsum[201]; strings; intmain() { while(cin>>s&&!(s.length()==1&&s[0]=='0')) { intx=s.length(); intcnt=x-1; while(cnt>=0) { sum[200-x+1+cnt]+=s[cnt]-4
OZTOET
·
2016-01-28 12:00
hdu1047 Integer
Inquiry
多次大数相加
转载请注明出处:http://blog.csdn.net/u012860063 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1047 Problem Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his
·
2015-11-13 14:41
Integer
POJ1503-Integer
Inquiry
转载请注明出处:優YoU http://user.qzone.qq.com/289065406/blog/1299062368 提示:就是多个大数相加的问题= = 1 //Memory Time 2 //184K 0MS 3 4 #include<iostream> 5 #include<cstring> 6 using name
·
2015-11-13 10:48
Integer
HDU 1047 Integer
Inquiry
http://acm.hdu.edu.cn/showproblem.php?pid=1047 大数 View Code import java.io.*; import java.math.*; import java.util.*; import java.text.*; public class Main { public static void main(S
·
2015-11-13 00:17
Integer
HDU 1047 Integer
Inquiry
题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1047 多个数的大数相加问题 View Code 1 #include<stdio.h> 2 #include<string.h> 3 int a[100000]; 4 char b[100000]; 5 int main() 6 { 7
·
2015-11-13 00:23
Integer
java写的大数算法
nbsp; 例题: n POJ 1503 Integer
Inquiry
·
2015-11-12 19:33
java
[投票]3·15消费者日 《程序员》杂志之现状调查
这就是这次调查的意义所在,没有奖品,只有真诚的交流,相关链接为:http://survey.csdn.net/
Inquiry
/ViewInquiryStep.asp
·
2015-11-12 16:36
程序员
Spoken Language Five
If you have urgent
inquiry
abo
·
2015-11-12 15:22
language
UVA 424 - Integer
Inquiry
很水的题。。。思路:这么长的数,首先要把他看成是字符串,存在一个字符数组里。(可不可以把它看成紧挨着的N数呢,每次读一个数直接进行加运算,最后的零没法处理,还没想到),让后把字符转换成数字存在整数数组里面,在里面进行运算。 #include<string.h>//#include<stdlib.h>#include<stdio.h>#define MAXN
·
2015-11-12 15:27
Integer
杭电 1047 Integer
Inquiry
import java.io.*; import java.util.*; import java.math.*; public class Main { public static void main(String args[]) { Scanner cin=new Scanner(System.in); int t;
·
2015-11-12 11:39
Integer
HDUOJ-----1074 Integer
Inquiry
Integer
Inquiry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768
·
2015-11-11 19:03
Integer
华丽的大数类,华丽的AC:Integer
Inquiry
1 /* UVa 424 - Integer
Inquiry
*/ 2 # include <iostream> 3 # include <string> 4 # include
·
2015-11-11 12:06
Integer
Integer
Inquiry
高精度加法,不知错在哪里。、 1 # include <stdio.h> 2 # include <string.h> 3 4 # define MAXN 110 5 6 char s[MAXN], sum[MAXN]; 7 8 int main() 9 {10 int i, len, c, tmp;11 char ch;12 13
·
2015-11-11 12:05
Integer
uva 424 Integer
Inquiry
#include<iostream> #include<cctype> #include<cstdlib> #include<cstring> using namespace std; int main(){ char a[110][110]; int result[110]; int num = 0,in
·
2015-11-11 11:57
Integer
hdu 1047 Integer
Inquiry
(高精度数)
Problem Description Oneof the first users of BIT's new supercomputer was Chip Diller. He extended hisexploration of powers of 3 to go from 0 to 333 and he explored taking varioussums of those number
·
2015-11-11 02:38
Integer
Poj 1503 Integer
Inquiry
id=1503 2.题目: Integer
Inquiry
Time Limit: 1000MS Memory Limit: 10000K Total Submissions
·
2015-11-10 22:53
Integer
HDU 1047 Integer
Inquiry
Integer
Inquiry
Problem Description One of the first users of BIT's
·
2015-11-08 11:33
Integer
[置顶] [Android源码分析]
inquiry
result引起的上层变化分析
8、
inquiry
result引起的上层变化  
·
2015-11-08 10:39
Android源码
Letter of
inquiry
about employment possibilities, e-mail version
Inquiry
about software engineering position after completion of M.S. in computer engineering December
·
2015-11-08 10:53
version
hdu 1047 Integer
Inquiry
pid=1047 Integer
Inquiry
Description One of the first users of BIT's new supercomputer was Chip Diller
·
2015-11-07 15:37
Integer
【大数加法】POJ-1503、NYOJ-103
1503:Integer
Inquiry
总时间限制: 1000ms 内存限制: 65536kB
·
2015-11-07 14:59
poj
POJ1503——字符串——Integer
Inquiry
Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.&nb
·
2015-11-07 14:05
Integer
UVa 424 Integer
Inquiry
【大数相加】
Integer
Inquiry
One of the firstusers of B
·
2015-11-07 14:36
Integer
POJ1503: Integer
Inquiry
(连续多个大整数加法运算)
1 #include<iostream> 2 #include<cstring> 3 using namespace std; 4 string sum; 5 const int max_len = 110; 6 string tool(string a){ 7 int dif = max_len - a.length(); 8
·
2015-11-05 08:58
Integer
Integer
Inquiry
--POJ 1503
1、题目类型:大数加法。 2、解题思路:水题。 3、注意事项:STL中reverse的巧用,加法进位的处理。 4、实现方法:(C++) #include < iostream > #include < string > #include < algorithm >
·
2015-11-02 16:34
Integer
HDOJ1047(Integer
Inquiry
)
Integer
Inquiry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768
·
2015-11-02 15:32
Integer
Integer
Inquiry
【大数的加法举例】
Integer
Inquiry
Time Limit: 1000MS Memory Limit: 10000K Total Submissions:
·
2015-11-02 13:31
Integer
建行“网站查询”系统非常不安全!
http://***/advisories/CCB_account_
inquiry
_security_problem.htm 笔者以前信用卡账户查询都是使用电话查询的,今日由于电话难打而去建行网站看看是否可以在线查询
·
2015-11-02 13:29
安全
UVa 424 Integer
Inquiry
之前杭电上也做过a + b的高精度的题,不过这道题的区别是有多组数据。 之前做的时候开了3个字符数组a,b,c,在计算的时候还要比较a,b长度,短的那个还要加'0',还设置了一个add来存放进位。 现在看来这种算法确实很繁琐。 而这次只用了两个字符数组,一个放加数,一个放和。 相比之前程序更短小了,而且可读性也提高了。 果然办法都是逼出来的。 没有了add,在判断进位的时候就看那一位的
·
2015-11-02 11:22
Integer
php如何获取本地手机号
php function
inquiry
_number_infor($phonenumber) /* *传入手机号码,通过API的到xml格式数据,对xml进一步解析,最后返回相应的号码信息的数组
·
2015-11-01 15:37
PHP
POJ 1503 Integer
Inquiry
/**/ /************************************** Problem: POJ 1503 Integer
Inquiry
·
2015-11-01 09:18
Integer
UVA 424 (13.08.02)
Integer
Inquiry
One of the first users of BIT's new supercomputer was Chip Diller.
·
2015-10-31 18:42
uva
35-2 Qualified Recommendation of a Former Employee
Situation: L.Richard Maxwell, Sales Manager for a national sporting goods manufacturer, has received an
inquiry
·
2015-10-31 13:24
form
UVA424 - Integer
Inquiry
题目链接:UVA424 大数相加问题,模拟手算即可,具体看代码: #include<iostream> #include<cstdio> #include<cstring> using namespace std; char number[255]={'0'}; int answer[260]={0}; void reverse(
·
2015-10-31 11:17
Integer
【原】 POJ 1503 Integer
Inquiry
大整数加法 解题报告
http://poj.org/problem?id=1503 方法: 很简单,模拟手动加法过程, 结果中的前置0要消去 Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of pow
·
2015-10-31 11:38
Integer
HDU 1047 Integer
Inquiry
(高精度加法)
Integer
Inquiry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768
·
2015-10-31 11:43
Integer
424 Integer
Inquiry
(整数查询)
Integer
Inquiry
One of the first users of BIT's new supercomputer
·
2015-10-31 10:16
Integer
Integer
Inquiry
_hdu_1047(大数).java
Integer
Inquiry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768
·
2015-10-31 10:11
Integer
交易策略 转
2012 | Views: 274 Economic Regulators According to Adam Smith's theory set forth in his book "An
Inquiry
·
2015-10-31 09:10
转
POJ 1503 Integer
Inquiry
大数 难度:0
题目链接:http://poj.org/problem?id=1503 1 import java.io.*; 2 import java.math.BigInteger; 3 import java.util.Scanner; 4 5 public class Main { 6 public static void main(Strin
·
2015-10-30 17:12
Integer
手机估值计算的jquery代码
<script type="text/javascript"> $('#
inquiry
').click(function(){
·
2015-10-30 14:57
jquery
424 - Integer
Inquiry
Integer
Inquiry
One of the first users of BIT's new supercomputer was Chip Diller.
·
2015-10-30 11:34
Integer
HDOJ 1047 Integer
Inquiry
Integer
Inquiry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768
·
2015-10-27 15:41
Integer
Account
Inquiry
(总账与子账的关联 - Cost Management - Receiving)
select pha.segment1 po_number, pv.vendor_name, pv.segment1 supplier_num, &n
·
2015-10-27 15:57
count
Account
Inquiry
(总账与子账的关联 - Payables - Purchase Invoices)
select aia.invoice_num, aila.line_number, aia.invoice_id, pha.segment1 po_number,  
·
2015-10-27 15:57
count
POJ 1503 Integer
Inquiry
简单大数相加
DescriptionOneofthefirstusersofBIT’snewsupercomputerwasChipDiller.Heextendedhisexplorationofpowersof3togofrom0to333andheexploredtakingvarioussumsofthosenumbers.Thissupercomputerisgreat,''remarkedChip.
qq_26525215
·
2015-10-24 13:00
poj
九度OJ 1119:Integer
Inquiry
(整数相加) (大数运算)
时间限制:1秒内存限制:32兆特殊判题:否提交:679解决:357题目描述: OneofthefirstusersofBIT'snewsupercomputerwasChipDiller. Heextendedhisexplorationofpowersof3togofrom0to333andheexploredtakingvarioussumsofthosenumbers. "Th
thudaliangrx
·
2015-10-23 23:00
C语言
大数运算
OJ
九度
poj 1503 Integer
Inquiry
(高精度运算)
题目链接:http://poj.org/problem?id=1503 思路分析: 基本的高精度问题,使用字符数组存储然后处理即可。 代码如下: #include <iostream> #include <string> using namespace std; const int M = 100 + 10; char inpu
·
2015-10-21 13:09
Integer
HDU_1047 Integer
Inquiry
(高精度)
在说这题之前请容我先说几句:1047!你妹是multiple test cases!你全家都是multiple test cases!!!丫,高精度没啥说的,光格式WA了两次!这不是浪费时间吗! #include <iostream>#include <cstdio>#include <cstring>using namespace
·
2015-10-21 11:41
Integer
A SCSI command code -- SIMPLIFIED DIRECT-ACCESS DEVICE (RBC)
B Description ------------------------------------------ 00 M TEST UNIT READY 12 M
INQUIRY
·
2015-10-21 10:07
command
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