蠻力法：

int maxSum1(int a[], int n){
    int i;
    int j;
    int maxSum = 0;

    for(i = 0; i < n; i++){
        int sum = 0;
        for(j = i; j < n; j++){
            sum += a[j];
            if(sum > maxSum){
                maxSum = sum;
            }
        }
    }


    return maxSum;

}
 

分治法：

int maxSum2(int left, int right, int a[]){
    int center , i;
    int sum, left_sum, right_sum;
    int left_max, right_max;

    if(left == right)
        return a[left] ;
    else{
            center = (left + right) / 2;
            left_sum = maxSum2(left, center, a);
            right_sum = maxSum2(center+1, right, a);
            sum = 0;
            left_max = 0;
            //求左邊的最大值
            for(i = center; i >= left; i--){
                sum += a[i];
                if(sum > left_max){
                    left_max = sum;
                }
            }
            sum = 0;
            right_max = 0;
            //求右邊的最大值
            for(i = center + 1; i <= right; i++){
                sum += a[i];
                if(sum > right_max){
                    right_max = sum;

                }
            }
            sum = right_max+ left_max;
            if(sum < left_sum)
                sum = left_sum;
            if(sum < right_sum)
                sum = right_sum;

    }
    return sum ;
}
 

動態規劃法：

void maxSum3(int a[], int n){
    int b[n];
    int i, j, s = n - 1;
    int maxSum = 0;
    b[0] = a[0]; //只有一個數時  最大子段和是其本身
    for(i = 1; i < n; i++){
        if(b[i - 1] > 0)
            b[i] = b[i - 1] + a[i];
        else{  //前面的值小於0  直接從當前原陣列開始
            b[i] = a[i];
        }
    }
 /*   printf("\n\n動態規劃法 列印b陣列:");
    for(i = 0; i < n; i++){
        printf("%d  ", b[i]);
    }
    printf("\n");
*/
    for(j = 0; j < n; j++){
        if(b[j] > maxSum){
            s = j;
            maxSum = b[j];

        }
    }
    printf("\n動態規劃法求出最大子段和為：%d", maxSum);

    for(i = s-1; i >= 0; i--){
        if(b[i] <= 0)
            break;
    }
    printf("\n\n列印最大子段：\n");
    for(i = i + 1; i <= s; i++){
        printf("%d  ", a[i]);
    }


}
 

完整程式碼：

#include <stdio.h>
#include <stdlib.h>
#define N 10
//判斷是否全為負數
int IsFuShu(int a[], int n){
    int i = 0;
    for(i = 0; i < n; i++){
        if(a[i] > 0)
            break;
    }
    if(i == n)
        return -1; //全是負數時
    else
        return 0; //有正數時


}

//蠻力法
//從第一個數開始，找第一個到最後一個數間最大欄位和；再從第二個數開始
int maxSum1(int a[], int n){
    int i;
    int j;
    int maxSum = 0;

    for(i = 0; i < n; i++){
        int sum = 0;
        for(j = i; j < n; j++){
            sum += a[j];
            if(sum > maxSum){
                maxSum = sum;
            }
        }
    }


    return maxSum;

}

//分治法
int maxSum2(int left, int right, int a[]){
    int center , i;
    int sum, left_sum, right_sum;
    int left_max, right_max;

    if(left == right)
        return a[left] ;
    else{
            center = (left + right) / 2;
            left_sum = maxSum2(left, center, a);
            right_sum = maxSum2(center+1, right, a);
            sum = 0;
            left_max = 0;
            //求左邊的最大值
            for(i = center; i >= left; i--){
                sum += a[i];
                if(sum > left_max){
                    left_max = sum;
                }
            }
            sum = 0;
            right_max = 0;
            //求右邊的最大值
            for(i = center + 1; i <= right; i++){
                sum += a[i];
                if(sum > right_max){
                    right_max = sum;

                }
            }
            sum = right_max+ left_max;
            if(sum < left_sum)
                sum = left_sum;
            if(sum < right_sum)
                sum = right_sum;

    }
    return sum ;
}

//動態
void maxSum3(int a[], int n){
    int b[n];
    int i, j, s = n - 1;
    int maxSum = 0;
    b[0] = a[0]; //只有一個數時  最大子段和是其本身
    for(i = 1; i < n; i++){
        if(b[i - 1] > 0)
            b[i] = b[i - 1] + a[i];
        else{  //前面的值小於0  直接從當前原陣列開始
            b[i] = a[i];
        }
    }
 /*   printf("\n\n動態規劃法 列印b陣列:");
    for(i = 0; i < n; i++){
        printf("%d  ", b[i]);
    }
    printf("\n");
*/
    for(j = 0; j < n; j++){
        if(b[j] > maxSum){
            s = j;
            maxSum = b[j];

        }
    }
    printf("\n動態規劃法求出最大子段和為：%d", maxSum);

    for(i = s-1; i >= 0; i--){
        if(b[i] <= 0)
            break;
    }
    printf("\n\n列印最大子段：\n");
    for(i = i + 1; i <= s; i++){
        printf("%d  ", a[i]);
    }


}

void Print(int a[], int n){
    printf("列印陣列:\n");
    int i;
    for(i = 0; i < n; i++){
        printf("%d   ", a[i]);
    }
}

void Creat(int a[]){
    int i,m;

    printf("輸入陣列的值：");
    for(i = 0; i < N; i++){
        scanf("%d", &m);
        a[i] = m;
    }

}
int main()
{
  //  int a[N] = {1,-1, 0, 2, -3, 4, -1, 0, -2, 4};
    int a[N];
    Creat(a);
    Print(a, N);
    int sum = IsFuShu(a, N);
    if(sum == -1)//全是負數值為0
        printf("最大子段和是0");
    else{
        sum = maxSum1(a, N);
        printf("\n蠻力法求出最大子段和是：%d\n" , sum);

        sum = maxSum2(0, N-1, a);
        printf("\n分治法求出最大子段和是：%d\n" , sum);

        maxSum3(a, N);


    }

    return 0;
}

執行結果：