转化N�q�制

明王不动 — Fri, 29 Aug 2008 08:34:00 GMT

#include <iostream>
#include <stack>
using namespace std;
int main()
{
int s;
int n;
int k;
cout<<"输入一个需要�{化进制的整数�Q?/span>"<<endl;
cin>>s;
cout<<"输入你想要�{化的�q�制�Q?�Q?6�Q?"<<endl;
cin>>n;
stack <int> stk;
do
{
stk.push( s % n);
s = s/n;
} while (s!=0);
while(!stk.empty())
{ k = stk.top();
stk.pop();
cout<< k<<" ";
}
return 0;
}

明王不动 2008-08-29 16:34 发表评论

find�q�算

明王不动 — Mon, 18 Aug 2008 08:36:00 GMT

1.find �q�算
find �q�算是基于�P代器的，因此可在��L��容器中��用相同的find函数查找倹{�?br>下面举些��单例子：(x��)
#include <iostream>
#include <algorithm>
#include <list>
#include <iterator>
using namespace std;
int main()
{
list <int> l;
int t;
while(cin>>t)
l.push_front(t);
int se;
cout<<"输入你想要查扄��?"<<endl;
cin.clear();
cin.sync();
cin>> se;
list<int> :: const_iterator result = find (l.begin

(),l.end(),se);
cout<<"the value "<<se<< (result == l.end()? " is not

present": "is present")<
return 0;
}
�c�M��圎ͼ��׃��指针的行��Z��作用在内�|�数�l�的�q�代器一��P��因此可用find来搜索数�l�：(x��)
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int a[] = {1,2,3,4,5};
int b = 7;
int *p = find(a,a+5,b);
cout<<"the value "<<b<<(p==a+5? "is not present":"is

present")<
return 0;
}

明王不动 2008-08-18 16:36 发表评论

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转化N�q�制

find�q�算