请使用c语言分治算法编写,给定一个有n(n≥1)个整数的序列,要求求出其中最大连续子序列的和。规定一个序列最大连续子序列和至少是0,如果小于0,其结果为0。求序列(-6,2,4,-7,5,3,2,-1,6,-9,10,-2)的最大子序列和为16
时间: 2024-05-03 09:20:48 浏览: 165
好的,我明白了您的要求。以下是使用C语言分治算法编写的代码:
```c
#include <stdio.h>
int max(int a, int b, int c) {
int max = a;
if (b > max) max = b;
if (c > max) max = c;
return max;
}
int maxCrossingSum(int arr[], int low, int mid, int high) {
int sum = 0;
int leftSum = -1000000; // 假设数组中最小值为-1000000,初始值为该值
for (int i = mid; i >= low; i--) {
sum += arr[i];
if (sum > leftSum) leftSum = sum;
}
sum = 0;
int rightSum = -1000000;
for (int i = mid+1; i <= high; i++) {
sum += arr[i];
if (sum > rightSum) rightSum = sum;
}
return max(leftSum+rightSum, leftSum, rightSum); // 跨越中间点的最大子序列和为三者中最大值
}
int maxSubArraySum(int arr[], int low, int high) {
if (low == high) return arr[low];
int mid = (low + high) / 2;
return max(maxSubArraySum(arr, low, mid),
maxSubArraySum(arr, mid+1, high),
maxCrossingSum(arr, low, mid, high));
}
int main() {
int arr[] = {-6, 2, 4, -7, 5, 3, 2, -1, 6, -9, 10, -2};
int n = sizeof(arr) / sizeof(arr[0]);
int maxSum = maxSubArraySum(arr, 0, n-1);
printf("The maximum sum of contiguous subsequence is %d", maxSum);
return 0;
}
```
输出结果为:
```
The maximum sum of contiguous subsequence is 16
```
这是使用分治算法求解最大连续子序列和的经典问题,算法的时间复杂度为O(nlogn)。具体实现过程请参考代码注释。
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