A-Anagram

連結：https://www.nowcoder.com/acm/contest/123/A
來源：牛客網
 

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

複製

ABCA BACA
ELLY KRIS
AAAA ZZZZ

輸出

複製

0
29
100

題目解析

          這道題是省賽的A題，題面比較長但是大體意思比較簡單。其實就是我們在第一個字串中依次取字元和第二個字串中的每個字元進行挨個的比較。選出最小的距離然後依次相加，最後得到一個距離和叫做sum並進行輸出。

           給定兩個字串，假設是ELLY與KRIS，E到K是6，L到R是6，當第二個L到I時，L是比I大的，此時L就要繞到Z，從Z到A，再從A開始到I，這樣長度就是23，Y到S同理，長度是20；這樣找完之後序列長度之和就是6 +6+23+20=55.這是題目中給出的一種解答。但是題目要求我們找字元間最小的長度，我就把第一個字串中的每一個字元與第二個字串中的每一個字元比較，每次都找出最短的長度，然後加在一起即可。

題目原始碼

#include<iostream>
#include<cstdio>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;
char str1[55],str2[55];
char s1[55],s2[55];
int vis[100];
const int maxn=1<<30;
int main(){
    while(~scanf("%s %s",str1,str2)){
        memset(vis,0,sizeof(vis));//將訪問陣列初始化為未訪問
        int sum=0;
        int len1 = strlen(str1);
        int len2 = strlen(str2);
        for(int i=0;i<len1;i++)
            s1[i]=str1[i];
        for(int i=0;i<len2;i++)
            s2[i]=str2[i];
        for(int i=0;i<len1;i++){
            int minn=maxn;
            int temp;
            for(int j=0;j<len2;j++){
               if(vis[j]==0){
                //如果未訪問過
                    if((s1[i]-'0')-(s2[j]-'0')>0){
                        //字串1的ascii大於字串2的ascii
                    if(abs((s1[i]-'0')-(s2[j]-'0')-26)<minn)
					   {
					    	minn=abs((s1[i]-'0')-(s2[j]-'0')-26);
						    temp=j;
					   }

                    }
                    else{
                        if(abs((s1[i]-'0')-(s2[j]-'0'))<minn){
                            minn=abs((s1[i]-'0')-(s2[j]-'0'));
                            temp=j;
                            printf("minn2=%d\n",minn);
                        }
                    }
               }
            }
            vis[temp]=1;
            sum+=minn;

        }
        printf("%d\n",sum);
    }
    return 0;
}