使用c语言:给定n个整数(可能为负数)组成的序列a[1],a[2],a[3],…,a[n],求该序列如a[i]+a[i+1]+…+a[j]的子段和的最大值。当所给的整数均为负数时定义子段和为0,依此定义,所求的最优值为: Max{0,a[i]+a[i+1]+…+a[j]},1<=i<=j<=n 例如,当(a[1],a[2],a[3],a[4],a[5],a[6])=(-20,11,-4,13,-5,-2)时,最大子段和为20(11,-4,13)。(使用分治算法分别实现)
时间: 2023-11-30 19:05:07 浏览: 189
MAX SUM 给定由n整数(可能为负数)组成的序列 {a1,a2,…,an},求该序列形如ai+ai+1,…,+aj的子段和的最大值。当所有的整数均为负数时定义其最大子段和为0。
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以下是使用分治算法实现求解最大子段和的C语言代码:
```c
#include <stdio.h>
#define INF -2147483647
// 求解跨越中点的最大子段和
int maxCrossingSum(int arr[], int left, int mid, int right) {
int leftSum = INF, rightSum = INF, sum = 0;
for (int i = mid; i >= left; i--) {
sum += arr[i];
if (sum > leftSum) {
leftSum = sum;
}
}
sum = 0;
for (int i = mid + 1; i <= right; i++) {
sum += arr[i];
if (sum > rightSum) {
rightSum = sum;
}
}
return leftSum + rightSum;
}
// 分治求解最大子段和
int maxSubarraySum(int arr[], int left, int right) {
if (left == right) {
return arr[left] > 0 ? arr[left] : 0;
}
int mid = (left + right) / 2;
int leftMaxSum = maxSubarraySum(arr, left, mid);
int rightMaxSum = maxSubarraySum(arr, mid + 1, right);
int crossMaxSum = maxCrossingSum(arr, left, mid, right);
int maxSum = leftMaxSum > rightMaxSum ? leftMaxSum : rightMaxSum;
maxSum = maxSum > crossMaxSum ? maxSum : crossMaxSum;
return maxSum;
}
int main() {
int arr[] = {-20, 11, -4, 13, -5, -2};
int n = sizeof(arr) / sizeof(arr[0]);
int maxSum = maxSubarraySum(arr, 0, n - 1);
printf("最大子段和为 %d\n", maxSum);
return 0;
}
```
运行结果为:
```
最大子段和为 20
```
该算法的时间复杂度为O(nlogn),比暴力算法的O(n^2)更优。
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