//求一个字符串中连续出现次数最多的子串
#include <iostream>
#include <string>
#include <vector>
using namespace std;
pair<int,string> fun(const string &str)
{
vector<string> substrs;
int maxcount=1,count =1;
string substr;
int i,len=str.length();
for(i=0; i <len; ++i)
substrs.push_back(str.substr(i,len-i));//把子串输入到容器对象substrs里面去
for(i=0; i <len ; ++i)
{
for (int j=i+1; j <len; ++j)
{
count =1;
if (substrs[i].substr(0,j-i) == substrs[j].substr(0,j-i))
{
++count;
for (int k=j+(j-i); k< len; k+=j-i)
{
if (substrs[i].substr(0,j-i)==substrs[k].substr(0,j-i))
++count;
else
break;
}
if (count > maxcount)
{
maxcount = count;
substr= substrs[i].substr(0,j-i);
}
}
}
}
return make_pair(maxcount,substr);
}
int main()
{
string str;
pair<int,string> rs;
while (cin>>str)
{
rs= fun(str);
cout << rs.second << ':' << rs.first << '\n';
}
return 0;
}
#include <iostream>
#include <string>
#include <vector>
using namespace std;
pair<int,string> fun(const string &str)
{
vector<string> substrs;
int maxcount=1,count =1;
string substr;
int i,len=str.length();
for(i=0; i <len; ++i)
substrs.push_back(str.substr(i,len-i));//把子串输入到容器对象substrs里面去
for(i=0; i <len ; ++i)
{
for (int j=i+1; j <len; ++j)
{
count =1;
if (substrs[i].substr(0,j-i) == substrs[j].substr(0,j-i))
{
++count;
for (int k=j+(j-i); k< len; k+=j-i)
{
if (substrs[i].substr(0,j-i)==substrs[k].substr(0,j-i))
++count;
else
break;
}
if (count > maxcount)
{
maxcount = count;
substr= substrs[i].substr(0,j-i);
}
}
}
}
return make_pair(maxcount,substr);
}
int main()
{
string str;
pair<int,string> rs;
while (cin>>str)
{
rs= fun(str);
cout << rs.second << ':' << rs.first << '\n';
}
return 0;
}
本文介绍了一种算法,用于找出一个字符串中连续出现次数最多的子串及其出现次数。通过遍历所有可能的子串并比较,最终确定出现频率最高的子串。
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