标签：stl oj acm string poj
Power Strings
| Time Limit: 3000MS |
|
Memory Limit: 65536K |
| Total Submissions: 30293 |
|
Accepted: 12631 |
Description
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the
empty string) and a^(n+1) = a*(a^n).
Input
Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.
Output
For each s you should print the largest n such that s = a^n for some string a.
Sample Input
abcd
aaaa
ababab
.
Sample Output
1
4
3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Source
题意:不多解释
这题书上说是用KMP来做，自己开始也尝试这样做，但后来发现根本就没有这个必要~~~几个简单的字符串处理函数就可以A掉了
代码:
#include<iostream>
#include<string>
using namespace std;
int main()
{
string s;
while(cin>>s&&s!=".")
{
int L=s.size();
for(int i=1;i<=L;i++)
{
int temp;
string str="";
if(L%i==0)
{
temp=L/i;
t=s.substr(0,i);
for(int j=0;j<temp;j++)
str+=t;
if(str==s)
{
cout<<temp<<endl;
break;
}
}
}
}
return 0;
}
POJ2406Power Strings,布布扣,bubuko.com
POJ2406Power Strings
标签：stl oj acm string poj
原文地址：http://blog.csdn.net/mummyding/article/details/25243369
