一道極其簡(jiǎn)單的ACM題,為什么第2段代碼通不過(guò)測(cè)試?
http://acm.hziee.edu.cn/showproblem.php?pid=1096
題目
Your task is to calculate the sum of some integers.

Input
Input contains an integer N in the first line, and then N lines follow. Each line starts with a integer M, and then M integers follow in the same line.

Output
For each group of input integers you should output their sum in one line, and you must note that there is a blank line between outputs.

Sample Input
3
4 1 2 3 4
5 1 2 3 4 5
3 1 2 3

Sample Output
10
15
6

第一段代碼,通過(guò)測(cè)試
# include <iostream>
using namespace std;
int main()
{
?int m,n,s;
?int sum;
?cin>>n;
?for(int i=0;i<n;i++)
?{
??if(i)
???cout<<endl;
??cin>>m;
??sum=0;
??for(int j=0;j<m;j++)
??{
???cin>>s;
???sum+=s;
??}
??cout<<sum<<endl;
?}
?return 0;
}

第2段代碼,通不過(guò)測(cè)試,why?
#include <iostream>
#include <vector>
#include <numeric>
#include <string>
#include <sstream>
using namespace std;
int main(void)
{
int times;
cin>>times;
while(times--!=-1){
??????? string str;
??????? istringstream istr;
??????? int temp;

??????? getline(cin,str);
??????? istr.str(str);
??????? vector<int> array;
???????
??????? bool ignore=true;
??????? while(!istr.eof()){
?????????? istr>>temp;??????????????????????????
?????????? if(ignore==true) {ignore=false;continue;}?????
?????????? array.push_back(temp);
?????????? }
??????? if(!array.empty()) cout<<accumulate(array.begin(), array.end(),0)<<"\n"<<endl;???
?}

return 0;
}

有時(shí)候第2段代碼會(huì)得出奇怪的結(jié)果,這是為什么?