HDU 3819 A and B Problem

A and B Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 786    Accepted Submission(s): 251

Problem Description

After calculating A + B, let’s consider another easy problem which only contains A and B.
You are given a string s containing only the letters 'A' and 'B'. The letters are arranged in a circle, so the last and first characters are adjacent. You will perform a series of swaps until all the 'A's form one consecutive sequence and all the 'B's form another consecutive sequence. In each swap, you can select any two characters and swap them. Find the minimal number of swaps necessary to reach your goal.

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case only contains a string as description.

Technical Specification

1. 1 <= T <= 100
2. 1 <= |S| <= 100000, |S| indicating the length of the string.

 

Output

For each test case, output the case number first, then the minimal number of swaps.

 

Sample Input

3AABBABABAABABA

 

Sample Output

Case 1: 0Case 2: 1Case 3: 1

 

Author

iSea@WHU

 

Source

The 6th Central China Invitational Programming Contest and 9th Wuhan University Programming Contest Final 

可以用循环数组也可以直接扩展成两倍计算。只需要选择一个范围，我们让这个范围大小是A的个数，一格一格的移动过去，判断所有情况中这个范围内B的个数最少是多少，这个就是需要移动的最少的次数了。

#include <vector>#include <map>#include <set>#include <algorithm>#include <iostream>#include <cstdio>#include <cmath>#include <cstdlib>#include <string>#include <cstring>using namespace std;string s,t;int main(){int n;scanf("%d",&n);int zl = 1;while(n--){cin >> s;int cnta = 0;int cnt = 0;for (int i = 0; i < s.size(); i++){if (s[i] == 'A') cnta++;}t = s + s;for (int i = 0; i < cnta; i++){if (t[i] == 'B') cnt++;}int end = cnta -1;int min_cnt = cnt;for (int i = 1; i < s.size(); i++){if (t[i - 1] == 'B') cnt--;if (t[end + 1] == 'B') cnt++;end++;min_cnt = min(cnt,min_cnt);}printf("Case %d: %d\n",zl++,min_cnt);}return 0;}