SRM147_DIV2

250

還是蠻簡單的

public class CCipher{
    public String decode(String cipherText, int shift){
        StringBuilder result = new StringBuilder();
        for (char i : cipherText.toCharArray()){
            int temp = i - 'A' - shift;
            if (temp < 0) temp += 26;
            result.append((char)(temp + 'A'));
        }
        return result.toString();

    }
}

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600

搞了將近一個小時，各種報錯。。

import java.util.Arrays;

public class PeopleCircle{
    public String order(int numMales, int numFemales, int K){
        int pointer = 0, allMen = numMales + numFemales;
        K--;
        int temp_k = K;
        char[] result = new char[allMen];
        Arrays.fill(result, 'M');
        if (numFemales == 0)
            return String.valueOf(result);
        while (numFemales > 0) {
            while (K > 0) {
                if (result[pointer] == 'M')
                    K--;
                pointer++;
                if (pointer >= allMen) {
                    pointer = pointer % allMen;
                }
            }
            K = temp_k;
            while (result[pointer] != 'M') {
                pointer++;
                if (pointer >= allMen) {
                    pointer = pointer % allMen;
                }
            }
                result[pointer] = 'F';
                pointer++;
                if (pointer >= allMen) {
                    pointer = pointer % allMen;
                }
                numFemales--;
            }

        return String.valueOf(result);
        }
}

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900

題本身不難，就是總是讀不懂題。大概就是把鏈子接在一起，需要砍斷鏈子中環的最小個數。如果一個鏈子的所有環都砍斷了，那麼會減少1次。就是儘可能的把鏈子拆了。所以就是先排序，在從最小的拆起。

import java.util.Arrays;

public class GoldenChain{
    public int minCuts(int[] sections){
        int cuts = 0;
        Arrays.sort(sections);
        for( int i = 0; i < sections.length; i++){
            if (sections.length - 1 - i >= sections[i] + cuts)cuts += sections[i];
            else{
                cuts += Math.max(0, sections.length - cuts -i);
                break;
            }
        }
        return cuts;
    }
}