看解答都是三维dp好像二维dp也能做dp[i][j]表示子串s[i...j]全变为1的最小操作数public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();String s = in.next();char[] cs = s.toCharArray();int res = 0;int[][] dp = new int[n][n];for(int i = 0; i < n; i++){dp[i][i] = cs[i] == '0' ? 1 : 0;res += cs[i] == '0' ? 1 : 0;}for (int i = n - 1; i >= 0; i--) {for(int j = i + 1; j < n; j++){if(cs[j] == '1'){dp[i][j] = dp[i][j - 1];}else{if(cs[j - 1] == '0'){dp[i][j] = dp[i][j - 1];}else{dp[i][j] = dp[i][j - 1] + 2;}}if((j - i + 1) % 2 != 0 &amp;&amp; dp[i][j] % 2 != 0){res ++;}}}System.out.println(res);}