時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 65536K，其他語言131072K
64bit IO Format: %lld

題目描述?

Orz has two strings of the same length:?A?and?B. Now she wants to transform?A?into an anagram of?B?(which means, a rearrangement of?B) by changing some of its letters. The only operation the girl can make is to “increase” some (possibly none or all) characters in?A. E.g., she can change an ‘A’ to a ‘B’, or a ‘K’ to an ‘L’. She can increase any character any times. E.g., she can increment an ‘A’ three times to get a ‘D’. The increment is cyclic: if she increases a ‘Z’, she gets an ‘A’ again.

????For example, she can transform “ELLY” to “KRIS” character by character by shifting ‘E’ to ‘K’ (6 operations), ‘L’ to ‘R’ (again 6 operations), the second ‘L’ to ‘I’ (23 operations, going from ‘Z’ to ‘A’ on the 15-th operation), and finally ‘Y’ to ‘S’ (20 operations, again cyclically going from ‘Z’ to ‘A’ on the 2-nd operation). The total number of operations would be?6 + 6 + 23 + 20 = 55. However, to make “ELLY” an anagram of “KRIS” it would be better to change it to “IRSK” with only 29 operations. You are given the strings?A?and?B. Find the minimal number of operations needed to transform A into some other string?X, such that?X?is an anagram of?B.

輸入描述:

There will be multiple test cases. For each testcase:There is two strings A and B in one line.∣A∣=∣B∣≤50. A and B will contain only uppercase letters from the English alphabet (‘A’-‘Z’).

輸出描述:

For each test case, output the minimal number of operations.示例1

輸入

ABCA BACA ELLY KRIS AAAA ZZZZ

輸出

0 29 100

題解：對輸入的兩個字符串先排序，如果a字符<=b的字符，改變的字符次數就是大的減小的，把改變成的字符，b標記為*表示改變了，繼續走，如果a>b字符，就用a-b+26,因為是環，所以加26。

#include<bits/stdc++.h> using namespace std; int main() {char a[100],b[100];int ans;while(~scanf("%s %s",a,b)){ans=0;int len=strlen(a);sort(a,a+len);sort(b,b+len);for(int i=0; i<len; i++)for(int j=0; j<len; j++){if(a[i]<=b[j]&b[j]!='*'){ans+=b[j]-a[i],b[j]='*';break;}if(j==len-1){for(int k=0; k<len; k++){if(b[k]!='*'){ans+=26-(a[i]-b[k]),b[k]='*';break;}}}}cout<<ans<<endl;}return 0; }

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