hdu1047 Integer Inquiry
/*
Integer Inquiry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15874 Accepted Submission(s): 4079 Problem 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.
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.) Input
The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative). The final input line will contain a single zero on a line by itself. Output
Your program should output the sum of the VeryLongIntegers given in the input. This problem contains multiple test cases! The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks. The output format consists of N output blocks. There is a blank line between output blocks. Sample Input 1 123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0 Sample Output 370370367037037036703703703670 Source
East Central North America 1996
*/
#include <iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxN=;
int main()
{
int n,k,i,j,len,next;
bool flag;
char ch[maxN],ans[maxN];
scanf("%d",&n);
while(n--)
{
flag=false;
memset(ans,,sizeof(ans));
for(i=; i<=maxN; i++)
{
scanf("%s",ch);
if(strcmp(ch,"")==)
break;
len=strlen(ch);
for(j=len-,k=; j>=; j--,k++)
{
next=;
ans[k]+=ch[j]-'';
while(ans[k+next]>)//must handle immediately
{
ans[k+next]-=;
ans[k++next]+=;
next+=;
}
}
}
for(j=maxN-; j>=; j--)//ans[] array can meet 0 advancely when you use strlen
{
if(ans[j]!='\0'&&!flag)
flag=true;
if(flag)
printf("%c",ans[j]+'');
}
if(!flag)//if nothing output before,we must output 0.
printf("%c",'');
printf("\n");
if(n>)
printf("\n");
}
return ;
}
hdu1047 Integer Inquiry的更多相关文章
- 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 ...
- (大数 string) Integer Inquiry hdu1047
Integer Inquiry Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) ...
- hdu acm-1047 Integer Inquiry(大数相加)
Integer Inquiry Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)T ...
- Integer Inquiry【大数的加法举例】
Integer Inquiry Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 27730 Accepted: 10764 ...
- 424 - Integer Inquiry
Integer Inquiry One of the first users of BIT's new supercomputer was Chip Diller. He extended his ...
- Integer Inquiry
Integer Inquiry Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Sub ...
- hdu 1047 Integer Inquiry
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1047 Integer Inquiry Description One of the first use ...
- 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 ...
- Poj 1503 Integer Inquiry
1.链接地址: http://poj.org/problem?id=1503 2.题目: Integer Inquiry Time Limit: 1000MS Memory Limit: 1000 ...
随机推荐
- 多余的Using Namespaces或引用会影响程序的执行效率么?
在.NET程序编写中,需要using相应命名空间或添加相应的References,可有时候没有使用到的命名空间也被添加到了Using Namespaces中,那么,这样会影响程序的执行效率么? 通过示 ...
- 当kfreebsd 用户遇见openSUSE系统
openSuse的系统工具集覆盖了四大主流桌面环境,是针对每一种桌面环境定制的独立的桌面体验.
- Follow me to learn what is repository pattern
Introduction Creating a generic repository pattern in an mvc application with entity framework is th ...
- ArcObject10.1降级至10.0
最开始接触ArcGIS版本是9.3,为了需要也安装了9.2进行开发:因为自己的电脑配置较低,所以跑不起10.0中文版:毕业工作后,行业内用10.1居多(虽然10.3已出):现在10.4都要出来了:由于 ...
- 关于SharePoint 的Client object model该何时load和execut query的一点自己的看法
很多人在用client object model的时候,不知道何时或者该不该load,今天看到一个观点描述这个问题,觉得很有道理,和大家分享.那就是写client object model就像写sql ...
- 如何在 在SharePoint 2013/2010 解决方案中添加 ashx (HttpHandler)
本文讲述如何在 在SharePoint 2013/2010 解决方案中添加 ashx (HttpHandler). 一般处理程序(HttpHandler)是·NET众多web组件的一种,ashx是其扩 ...
- android 保存 用户名和密码 设置等应用信息优化
1.传统的保存用户名,密码方式 SharedPreferences Editor editor = shareReference.edit(); editor.putString(KEY_NAME,& ...
- android:#FFFFFFFF 颜色码解析
原文地址:android:#FFFFFFFF 颜色作者:android小鸟 颜色色码为#FFFFFFFF 其中颜色顺序依次为#AARRGGBB 前两位AA代表透明度,FF时表示不透明,00表示透明: ...
- ubuntu解决arm-linux-gcc no such file的问题
这种情况是因为你的操作系统是Ubuntu 64位的,而交叉编译工具链都是32位执行程序.要成功运行这些交叉编译工具链,需要与这些工具链相关的32位库.安装命令如下:sudo apt-get insta ...
- Objective-C的IO流