Codeforces 145A-Lucky Conversion(规律)

A. Lucky Conversion

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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上的题确实质量很好，这题还是A题就卡了好一阵，思路很诡异。。

题意：
     给出一串字符串，只包含数字4和7 然后再给出另一个字符串，同样是只包含4和7，长度相同，现在给定两种操作，①：改变a串某位上的数字；②：交换a串任意两位上的数字，求最小操作数 使得a==b

因为是要最小操作数，所以要尽可能的执行交换操作，所以 扫一遍a串，看它和b串的对应位置上的数字是不是相同，若不同，记下4和7不同的个数，其中最大数就是答案。

#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\n",max(cnt1,cnt2));
    }
    return 0;
}