vector<int> v;
int num;
while (cin >> num)
{
v.push_back(num);
if (num == 0)
break;
}
方式一:以空格或者回车结束每次输入,以固定输入数字0结束输入(也可以更改为其他字符)
string test;
getline(cin, test);
int input_nums = 0;
vector<int> indexes;
indexes.push_back(0);
for (int i = 0; i < test.size(); i++)
{
if (test[i] == ' ')
{
input_nums++;
indexes.push_back(i);
}
}
int *p = new int[input_nums+1];
for (int i = 0; i <= input_nums; i++)
{
int start = indexes[i];
int length = (i!=input_nums?indexes[i + 1] - indexes[i]:test.length()- indexes[i]);
p[i] = atoi(test.substr(start, length).c_str());
}
方式二:以空格结束每次输入,将输入当成字符串来处理,以空格分隔字符并转换成int。相对复杂一点。。。