UVA - 424 Integer Inquiry

Integer Inquiry

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

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.

Sample Input

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

分析：

题意是计算各长整数的和，最多输入100个数，每个数都是正整数且位数不超过100.


很明显100位的数用数组存会方便许多，然后设置两个整型数组，一个存进位，一个存结果。为了对齐进行运算，我将所有的数先逆序调整，即低位在前高位在后，这样运算就能保证对齐运算了。总体来说难度很小，需要细心，争取一次AC！

代码：

#include<iostream>
#include<cstring>
using namespace std;
char input[101][101];
int convert(int size)
{
int maxn=0;
for (int i=0;i<size;i++)
{
int len=strlen(input[i]);
if (len>maxn) maxn=len;
for (int j=0;j<len/2;j++)
{
char tmp=input[i][j];
input[i][j]=input[i][len-1-j];
input[i][len-1-j]=tmp;
}
}
return maxn;
}
int main()
{
int line=0;
while(cin>>input[line]&&input[line][0]!='0')
{
line++;
}
int carry[102]={0};
int maxsize=convert(line);
int result[105];
for (int i=0;i<maxsize;i++)
{
int sum=0;
for (int j=0;j<line;j++)
{
if (input[j][i]=='\0') input[j][i]='0';
sum=sum+int(input[j][i]-'0');
}
sum=sum+carry[i];
result[i]=sum%10;
carry[i+1]=sum/10;
}
result[maxsize]=carry[maxsize];

for (int j=0;j<(maxsize+1)/2;j++)
{
int tmp=result[j];
result[j]=result[maxsize-j];
result[maxsize-j]=tmp;
}
for (int i=0;i<=maxsize;i++)
{
if (i==0&&result[i]==0) continue;
cout<<result[i];
}
cout<<endl;
return 0;
}