Codeforces 145A-Lucky Conversion(规律)

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 ainto string b.

Sample test(s)

input
47
74

output
1

input
774
744

output
1

input
777
444

output
3   不得不承认CF上的题确实质量非常好，这题还是A题就卡了好一阵，思路非常诡异。。题意：
     给出一串字符串，仅仅包括数字4和7 然后再给出还有一个字符串，相同是仅仅包括4和7，长度相同，如今给定两种操作，①：改变a串某位上的数字；②：交换a串随意两位上的数字，求最小操作数 使得a==b由于是要最小操作数，所以要尽可能的运行交换操作，所以 扫一遍a串，看它和b串的相应位置上的数字是不是同样，若不同，记下4和7不同的个数，当中最大数就是答案。#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
#define LL long long
const int maxn=100050;
char s[maxn],t[maxn];
int main()
{
    while(~scanf("%s%s",s,t))
    {
        int len=strlen(s),cnt1=0,cnt2=0;
        for(int i=0;i<len;i++)
        {
            if(s[i]!=t[i])
            {
                if(s[i]=='4')
                    cnt1++;
                else
                    cnt2++;
            }
        }
        printf("%d
",max(cnt1,cnt2));
    }
    return 0;
}