如何通过巧妙变换实现字符串的hdu 4357 String change？
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本文共计649个文字，预计阅读时间需要3分钟。
字符串+时间限制：2000/1000毫秒（Java/其他）内存限制：65536/32768K（Java/其他）总提交次数：802接受提交次数：370题目描述：在这个问题中，你将接收到两个只包含小写字母的字符串S1和S2。
String change
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 802Accepted Submission(s): 370
Problem Description
In this problem you will receive two strings S 1and S 2that contain only lowercase letters. Each time you can swap any two characters of S 1. After swap,both of the two letters will increase their value by one. If the previous letter is 'z',it will become 'a' after being swapped. That is to say ,"a" becomes "b","b" becomes "c"....."z" becomes "a" and so on. You can do the change operation in S 1as many times as you want. Please tell us whether you can change S 1to S 2after some operations or not.
Input
There are several cases.The first line of the input is a single integer T (T <= 41) which is the number of test cases.Then comes the T test cases . For each case,the first line is S 1,the second line is S 2.S 1has the same length as S 2and the length of the string is between 2 and 60.
Output
For each case,output "Case #X: " first, X is the case number starting from 1.If it is possible change S 1to S 2output "YES",otherwise output "NO".
Sample Input
3abbabacddbaaabbcbccd
Sample Output
Hint
Author
miketc@UESTC_Goldfinger
Source
2012 Multi-University Training Contest 6
Recommend
zhuyuanchen520
思路：推理后发现超过两张牌的情况，只要是奇偶一样都可以用变换。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int mm=99;char s[mm],t[mm];int cas;int main(){ while(~scanf("%d",&cas)) { for(int ca=1;ca<=cas;++ca) { scanf("%s%s",s,t); printf("Case #%d: ",ca); int len=strlen(s); bool a=0,b=0; bool flag=0; int k1,k2; if(len==2) { k1=(s[0]-t[0]+26)%26; k2=(s[1]-t[1]+26)%26; if(k1==k2&&k1%2==0)flag=1; k1=(s[0]-t[1]+26)%26; k2=(s[1]-t[0]+26)%26; if(k1==k2&&k1%2==1)flag=1; if(flag)printf("YES\n"); else printf("NO\n"); } else { for(int i=0;i<len;++i) { a^=s[i]&1; b^=t[i]&1; } if(a==b)printf("YES\n"); else printf("NO\n"); } } } return 0;}本文共计649个文字，预计阅读时间需要3分钟。
字符串+时间限制：2000/1000毫秒（Java/其他）内存限制：65536/32768K（Java/其他）总提交次数：802接受提交次数：370题目描述：在这个问题中，你将接收到两个只包含小写字母的字符串S1和S2。
String change
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 802Accepted Submission(s): 370
Problem Description
In this problem you will receive two strings S 1and S 2that contain only lowercase letters. Each time you can swap any two characters of S 1. After swap,both of the two letters will increase their value by one. If the previous letter is 'z',it will become 'a' after being swapped. That is to say ,"a" becomes "b","b" becomes "c"....."z" becomes "a" and so on. You can do the change operation in S 1as many times as you want. Please tell us whether you can change S 1to S 2after some operations or not.
Input
There are several cases.The first line of the input is a single integer T (T <= 41) which is the number of test cases.Then comes the T test cases . For each case,the first line is S 1,the second line is S 2.S 1has the same length as S 2and the length of the string is between 2 and 60.
Output
For each case,output "Case #X: " first, X is the case number starting from 1.If it is possible change S 1to S 2output "YES",otherwise output "NO".
Sample Input
3abbabacddbaaabbcbccd
Sample Output
Hint
Author
miketc@UESTC_Goldfinger
Source
2012 Multi-University Training Contest 6
Recommend
zhuyuanchen520
思路：推理后发现超过两张牌的情况，只要是奇偶一样都可以用变换。
#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int mm=99;char s[mm],t[mm];int cas;int main(){ while(~scanf("%d",&cas)) { for(int ca=1;ca<=cas;++ca) { scanf("%s%s",s,t); printf("Case #%d: ",ca); int len=strlen(s); bool a=0,b=0; bool flag=0; int k1,k2; if(len==2) { k1=(s[0]-t[0]+26)%26; k2=(s[1]-t[1]+26)%26; if(k1==k2&&k1%2==0)flag=1; k1=(s[0]-t[1]+26)%26; k2=(s[1]-t[0]+26)%26; if(k1==k2&&k1%2==1)flag=1; if(flag)printf("YES\n"); else printf("NO\n"); } else { for(int i=0;i<len;++i) { a^=s[i]&1; b^=t[i]&1; } if(a==b)printf("YES\n"); else printf("NO\n"); } } } return 0;}