SRM147_DIV2

250

还是蛮简单的

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();

}
}

600

搞了将近一个小时，各种报错。。

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);
}
}

900

题本身不难，就是总是读不懂题。大概就是把链子接在一起，需要砍断链子中环的最小个数。如果一个链子的所有环都砍断了，那么会减少1次。就是尽可能的把链子拆了。所以就是先排序，在从最小的拆起。

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;
}
}