题目链接:http://cpp.zjut.edu.cn/ShowProblem.aspx?ShowID=1885
题面:
Careless Tony
Time Limit:1000MS Memory Limit:32768K
2 1 WishingBone WashingBone 1 Oops Oooops
20 6
题意:
最少用多少时间修改后面那个串到前面那个串。
代码:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int n,t;
string s1,s2;
bool flag;
int p;
cin>>n;
while(n--)
{
cin>>t;
cin>>s1>>s2;
flag=true;
p=0;
for(int i=0;i<s1.length()&&i<s2.length();i++)
{
if(s1[i]!=s2[i])
{
flag=false;
p=i;
break;
}
}
if(flag)
{
if(s1.length()>=s2.length())
{
cout<<t*(s1.length()-s2.length())<<endl;
}
else
{
cout<<(s2.length()-s1.length())*t<<endl;
}
}
else
{
cout<<(s1.length()+s2.length()-2*p)*t<<endl;
}
}
return 0;
}原文地址:http://blog.csdn.net/david_jett/article/details/45104103