HDoj Integer Inquiry(大数)

本文主要是介绍HDoj Integer Inquiry(大数)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

真心要哭了。。这几天在搞大数高精度计算昨晚在机房敲很快敲完了就是过不了啊过不了劳资都想骂脏话啊 NMB 一开始不输出前面的0啊过不了看discuss 百度找了个AC的代码找了几组测试数据那个代码输出前面的0啊我的妈今天有找了个代码不输出0啊我的天。。。真心要被逼疯了幸好还是AC了。。。算是有进步吧之前的心态肯定坚持不下来啊

杭电ACM 2014暑期集训队——选拔安排~

Integer Inquiry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11714 Accepted Submission(s): 2942

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

      1123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

      370370367037037036703703703670

Source

East Central North America 1996

Recommend

题意很简单就是输入一连串数字作为一个大的整数遇到单个0结束输出

每个CASE输出之间输出一个空行

我的测试数据：

5
123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

000
00
0

02
088
0

9999
99
0

00
00
0

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
using namespace std;

const int maxn=110;
char s[maxn];
int res[maxn];
int num[maxn];

int main()
{
freopen("test.txt","r",stdin);
int n;
cin>>n;
getchar();
while(n--)
{
memset(res,0,sizeof(res));
int m=0;
while(1)
{
gets(s);
if(strlen(s)==1&&s[0]=='0')
break;
int k=0;
for(int i=strlen(s)-1;i>=0;i--)
{
num[k++]=s[i]-'0';
//cout<<num[k-1]<<" ";
}
//cout<<endl;
m=max(m,k);
//cout<<"m1="<<m<<endl;
if(m>k)
{
for(int i=k;i<m;i++)
num[i]=0;
}
for(int i=0;i<m;i++)
{
res[i]+=num[i];
res[i+1]+=res[i]/10;
res[i]=res[i]%10;
}

//cout<<"res[m]="<<res[m]<<"res[m-1]"<<res[m-1]<<endl;
if(res[m]>0)
{
res[m]+=res[m-1]/10;
res[m-1]=res[m-1]%10;
m++;
}
}
//cout<<"m2="<<m<<endl;
// for(int i=m-1;i>=0;i--)
// cout<<res[i];
int flag=0;
int x=m-1;
for(;x>=0;x--)
{
if(res[x]>0)
flag=1;
if(flag)
cout<<res[x];
}
if(flag==0)
cout<<0;//如果输入的全是0 输出0
cout<<endl;
if(n)
cout<<endl;
}
return 0;
}