给出一个只包含 0 和 1 的 01 串 s ，下标从 1 开始，设第 i 位的价值为 vali ，则价值定义如下：
1. i=1时:val1 = 1
2. i>1时：
2.1 若 si ≠ si-1 , vali = 1
2.2 若 si = si-1 , vali = vali-1 + 1
字符串的价值等于 val1 + val2 + val3 + ... + valn
[编程题]01串的价值
- 热度指数：3199 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 256M，其他语言512M
- 算法知识视频讲解
输入描述:
第一行一个正整数 n ，代表串长度。接下来一行一个 01 串 s 。1 ≤ n ≤ 5,000
输出描述:
输出一个整数代表答案
示例1
输入
6 010101
输出
7
说明
删除后的串为0001或0111时有最大价值
示例2
输入
20 11111000111011101100
输出
94
示例3
输入
4 1100
输出
6
import java.util.Scanner;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int original = in.nextInt();
String originalStr = in.next();
String currentStr = getCurrentStr(originalStr);
int length = currentStr.length();
int[] currentArrays = new int[length];
fillArrays(currentStr, currentArrays);
System.out.println(sumVal(length, currentArrays));
}
public static void fillArrays(String currentStr, int[] currentArrays) {
char[] chars = currentStr.toCharArray();
for (int i = 0; i < currentArrays.length; i++) {
currentArrays[i] = chars[i];
}
}
public static String getCurrentStr(String originalStr) {
int count1 = 0;
int count0 = 0;
char[] chars = originalStr.toCharArray();
for (int i = 0; i < chars.length; i++) {
if ('1' == chars[i]) {
count1++;
} else {
count0++;
}
}
if (count0 >= count1) {
originalStr = removeStr('1', '0', originalStr);
} else {
originalStr = removeStr('0', '1', originalStr);
}
return originalStr;
}
public static String removeStr(char removed, char reserved,
String originalStr) {
int firstIndex = originalStr.indexOf(reserved);
String firstStr = "";
if (firstIndex != 0) {
firstStr = originalStr.substring(0, firstIndex);
}
int lastIndex = originalStr.lastIndexOf(reserved);
String secondStr = originalStr.substring(firstIndex, lastIndex + 1);
String thirdStr = "";
if (lastIndex != originalStr.length() - 1) {
thirdStr = originalStr.substring(lastIndex + 1);
}
secondStr = secondStr.replaceAll(String.valueOf(removed), "");
originalStr = new StringBuilder().append(firstStr).append(secondStr).append(
thirdStr).toString();
return originalStr;
}
public static int sumVal(int length, int[] arrays) {
int sum = 0;
for (int i = 1; i < length + 1; i++) {
sum += getVal(i, arrays);
}
return sum;
}
public static int getVal(int i, int[] arrays) {
if (i == 1) {
return 1;
}
if (arrays[i - 1] != arrays[i - 2]) {
return 1;
}
return getVal(i - 1, arrays) + 1;
}
}
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int original = in.nextInt();
String originalStr = in.next();
String currentStr = getCurrentStr(originalStr);
int length = currentStr.length();
int[] currentArrays = new int[length];
fillArrays(currentStr, currentArrays);
System.out.println(sumVal(length, currentArrays));
}
public static void fillArrays(String currentStr, int[] currentArrays) {
char[] chars = currentStr.toCharArray();
for (int i = 0; i < currentArrays.length; i++) {
currentArrays[i] = chars[i];
}
}
public static String getCurrentStr(String originalStr) {
int count1 = 0;
int count0 = 0;
char[] chars = originalStr.toCharArray();
for (int i = 0; i < chars.length; i++) {
if ('1' == chars[i]) {
count1++;
} else {
count0++;
}
}
if (count0 >= count1) {
originalStr = removeStr('1', '0', originalStr);
} else {
originalStr = removeStr('0', '1', originalStr);
}
return originalStr;
}
public static String removeStr(char removed, char reserved,
String originalStr) {
int firstIndex = originalStr.indexOf(reserved);
String firstStr = "";
if (firstIndex != 0) {
firstStr = originalStr.substring(0, firstIndex);
}
int lastIndex = originalStr.lastIndexOf(reserved);
String secondStr = originalStr.substring(firstIndex, lastIndex + 1);
String thirdStr = "";
if (lastIndex != originalStr.length() - 1) {
thirdStr = originalStr.substring(lastIndex + 1);
}
secondStr = secondStr.replaceAll(String.valueOf(removed), "");
originalStr = new StringBuilder().append(firstStr).append(secondStr).append(
thirdStr).toString();
return originalStr;
}
public static int sumVal(int length, int[] arrays) {
int sum = 0;
for (int i = 1; i < length + 1; i++) {
sum += getVal(i, arrays);
}
return sum;
}
public static int getVal(int i, int[] arrays) {
if (i == 1) {
return 1;
}
if (arrays[i - 1] != arrays[i - 2]) {
return 1;
}
return getVal(i - 1, arrays) + 1;
}
}
发表于 2023-07-07 08:28:58
回复(0)
放个java版本的动态规划。dp[i][j][0] 表示到了第i位，上一位的是0，积累的价值为j，dp[i][j][1]表示到了第i位，上一位是1，积累的价值为j。计算状态转移方程，最后打印的答案是最后一位所有积累价值的最大值，答案测试案例确实有问题，别的楼层也提到了，可以自行检验。
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
in.nextLine();
String s = in.nextLine();
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = s.charAt(i) - '0';
}
int[][][] dp = new int[n][n + 1][2];
dp[0][1][array[0]] = 1;
for (int i = 1; i < n; i++) {
for (int j = 1; j <= n; j++) {
if (j == 1) {
for (int k = 0; k < n; k++) {
if (array[i] == 0) {
dp[i][1][0] = Math.max(dp[i][1][0], dp[i - 1][k][1] + 1);
dp[i][1][1] = dp[i - 1][1][1];
} else {
dp[i][1][0] = dp[i - 1][1][0];
dp[i][1][1] = Math.max(dp[i][1][1], dp[i - 1][k][0] + 1);
}
}
} else {
if (array[i] == 0) {
if (dp[i - 1][j - 1][0] != 0) {
dp[i][j][0] = dp[i - 1][j - 1][0] + j;
}
dp[i][j][1] = dp[i - 1][j][1];
} else {
dp[i][j][0] = dp[i - 1][j][0];
if (dp[i - 1][j - 1][1] != 0) {
dp[i][j][1] = dp[i - 1][j - 1][1] + j;
}
}
}
}
}
int result = 0;
for (int i = 0; i <= n; i++) {
result = Math.max(result, dp[n - 1][i][0]);
result = Math.max(result, dp[n - 1][i][1]);
}
System.out.println(result);
}
}
发表于 2023-03-26 09:21:11
回复(0)
//贪心
public int valueOf01String(String str) {
char[] s = str.toCharArray();
int n = s.length;
int a = -1, b = -1; //记录第一个0和第一个1
int c = -1, d = -1; //记录最后一个0和最后一个1
int cnt0 = 0, cnt1 = 0;//记录0 和1的个数
for(int i = 0; i < n; i++) {
if (a == -1 && s[i] == '0') a = i;
if (b == -1 && s[i] == '1') b = i;
if (s[i] == '0' && i > c) c = i;
if (s[i] == '1' && i > d) d = i;
if (s[i] == '0') cnt0++;
else cnt1++;
}
return Math.max(C(cnt0) + C(a) + C(n - c - 1), C(cnt1) + C(b) + C(n - d - 1));
}
int C(int n){return n * (n + 1) / 2;}
发表于 2023-03-07 20:00:59
回复(0)
问题信息
通过挑战的用户
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2023-03-14 16:52:50 -
2023-03-07 17:07:47
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2023-03-05 22:45:31
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2023-01-02 00:21:36
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2022-11-26 16:37:48
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