我需要翻译这个方程:
n = 7
1 + 1 - 4 - 4 - 4 - 2 - 2
我该如何最佳地替换运算符,使得方程式的总和等于零,或者打印
-1
。我想到了一种算法,但它不是最优解。我有一个想法,用
O(n*2^n)
的复杂度暴力破解所有情况,但(n<300)
。这是问题的链接: http://codeforces.com/gym/100989/problem/M。
我需要翻译这个方程:
n = 7
1 + 1 - 4 - 4 - 4 - 2 - 2
-1
。O(n*2^n)
的复杂度暴力破解所有情况,但(n<300)
。这里有一个想法:
在Java中实现:
// assuming the numbers are positive
// (ignore operator when parsing, line.split("[ +-]+") will do)
public static int canReach0(int[] numbers) {
sort(numbers, 1); // sort(array, offset) doesn't matter what algorithm
// for 300 elements and compared to the complexity of the rest
int[] revSum = new int[numbers.length];
revSum[numbers.length - 1] = numbers[numbers.length - 1];
for (int i = numbers.length - 2; i >= 0; i--)
revSum[i] = revSum[i + 1] + numbers[i];
int sum = numbers[0];
if (sum == revSum[1])
return 0;
return solve(numbers, 1, sum, revSum);
}
private static int solve(int[] numbers, int index, int sum, int[] revSum) {
if (index == numbers.length - 1)
return -1;
int high = sum + numbers[index];
if (high == revSum[index + 1] ||
(high < revSum[index + 1] && solve(numbers, index + 1, high, revSum) == 0))
return 0;
int low = sum - numbers[index];
if (low == -revSum[index + 1] ||
(low > -revSum[index + 1] && solve(numbers, index + 1, low, revSum) == 0))
return 0;
return -1;
}