这个题主要解决的是精度问题，若不同前面用0补上，字符串前加"0".
Integer Inquiry
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 7 Accepted Submission(s) : 2
Font: Times New Roman | Verdana | Georgia
Font Size: ← →
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
这个题主要解决的是：首先是把输入的数的位数相同，若不同前面补0
#include<stdio.h>
#include<string>
#include<iostream>
using namespace std;
//高精度加法
//只能是两个正数相加
string add(string str1,string str2)//高精度加法
{
string str;
int len1=str1.length();
int len2=str2.length();
//前面补0，弄成长度相同
if(len1<len2)
{
for(int i=1;i<=len2-len1;i++)
str1="0"+str1;
}
else
{
for(int i=1;i<=len1-len2;i++)
str2="0"+str2;
}
len1=str1.length();
int cf=0;
int temp;
for(int i=len1-1;i>=0;i--)
{
temp=str1[i]-‘0‘+str2[i]-‘0‘+cf;
cf=temp/10;
temp%=10;
str=char(temp+‘0‘)+str;
}
if(cf!=0) str=char(cf+‘0‘)+str;
return str;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
string sum="0";
string str1;
while(cin>>str1)
{
if(str1=="0")break;
sum=add(sum,str1);
}
cout<<sum<<endl;
if(T>0)cout<<endl;
}
return 0;
}