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
類似的可以看這篇博客:https://blog.csdn.net/weixin_43823808/article/details/98449076
程序代碼:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
#define N 101
int num[N],sum[N];
char s[N];
void find(char s[])
{
int len=strlen(s),j=0;
memset(num,0,sizeof(num));
for(int i=len-1;i>=0;i--)
num[j++]=s[i]-'0';
for(int i=0;i<N;i++)
{
sum[i]+=num[i];
if(sum[i]>=10)
{
sum[i]-=10;
sum[i+1]++;
}
}
}
int main()
{
int n;
cin>>n;
while(n--)
{
memset(sum,0,sizeof(sum));
while(cin>>s)
{
if(strcmp(s,"0")==0)
break;
find(s);
}
bool flag=false;
for(int i=N;i>=0;i--)
{
if(flag)
cout<<sum[i];
else if(sum[i])
{
cout<<sum[i];
flag=true;
}
}
if(!flag)
cout<<"0";
cout<<endl;
if(n)
cout<<endl;
}
return 0;
}