Problem: count how many 0's in a string consisting of 0 and 1 only

实在想不明白如何才能在不遍历整个串之前得知0的个数，了不起只能判断1的个数，当1的个数到达n-n/2-1时停止遍历，得出0的个数。但是意义也不大。
所以还请你明示！

may be you can change the string type to the bitset type and then you could transfer the operation of the bitset type ,just like that:
size_t num=bitvec.cout()//the bitvec is a object of bitset type and the number means that the 1's units;

another way :you can make a use of the iterator to count the 1's units in you string ,for exemple :

#include<iostream>
#include<string>
#include<cstddef>
using namespace std;
int main(){
size_t num0=0,num1=0;
string str("0001111000");
string::iterator iter;
for(iter=str.begin();iter!=str.end();++iter){
if(*iter!=' '){
if(*iter=='1')
num1++;
if(*iter=='0')
num0++;
}
else cout<<"errer!"<<endl;
}
cout<<num1<<endl;
cout<<num0<<endl;
return 0;
}
if you have some problems please show me and give me a reason ,thank you.
by the way ,i see ,i can't make sure that i understand you real idea.