CodeForces 145A

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本文介绍了一个简单的编程问题：如何通过最少的操作将一个由幸运数字组成的字符串转换为另一个相同长度的字符串。文中提供了一段示例代码，展示了如何通过比较两个字符串并计算所需的替换次数来解决该问题。

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CodeForces 145A

Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya has two strings a and b of the same length n. The strings consist only of lucky digits. Petya can perform operations of two types:

replace any one digit from string a by its opposite (i.e., replace 4 by 7 and 7 by 4);
swap any pair of digits in string a.

Petya is interested in the minimum number of operations that are needed to make string a equal to string b. Help him with the task.

Input

The first and the second line contains strings a and b, correspondingly. Strings a and b have equal lengths and contain only lucky digits. The strings are not empty, their length does not exceed 10⁵.

Output

Print on the single line the single number — the minimum number of operations needed to convert string a into string b.

Sample Input

Input

47
74

Output

Input

774
744

Output

Input

777
444

Output

Sample Output

Hint

In the first sample it is enough simply to swap the first and the second digit.

In the second sample we should replace the second digit with its opposite.

In the third number we should replace all three digits with their opposites.

同样的水题........

在第一个样品是足够简单的交换第一和第二位。

在第二个样本应与其相反的取代二位数。

在第三号应与他们对立取代所有的三位数。

代码也就这样了

#include<iostream>
#include<cstring>
using namespace std;
char a[100010],b[100010];
int main()
{
    int flag1 = 0;
    int flag2 = 0;
    int i;
    cin>>a;
    cin>>b;
    for(i=0; a[i]!='\0'; i++)
    {
        if(a[i]!=b[i])
        {
            if(a[i]=='7')
            {
                flag1++;
            }
            if(a[i]=='4')
            {
                flag2++;
            }
        }
    }
    if(flag1>flag2)//找出最大的就OK了
        cout<<flag1<<endl;
    else
        cout<<flag2<<endl;
    return 0;
}