Question

一些上下文：我正在解决HackerRank中关于连续子序列和一系列整数上具有最大和的子集的问题。该算法可能是正确的，我不是对它的正确性有所保留，而是我程序的异常行为，请继续阅读。

input.txt中：

1
6
2 -1 2 3 4 -5

输入格式：

number of testcases
number of integers (let that be n)
n space-sepearated integers

正确输出（基于来源）：

10 11

实际输出：

11 11

如您所见，实际输出与正确的输出不匹配。我花时间在编写程序之前和之后使用笔和纸手动执行输入算法，我的计算似乎与正确的输出匹配（我的算法可能仍然不正确）。但之后，我尝试使用gdb逐行跟踪程序，并且在函数调用bestSum1的开始时发现了一些异常。

GDB输出代码段 :(转到我打印整数[0]的末尾附近的部分，并显示当我将其传递给函数时值已更改。

(gdb) break main
Breakpoint 1 at 0x4005fe: file maxsub.c, line 14.
(gdb) break bestSum1
Breakpoint 2 at 0x400745: file maxsub.c, line 33.
(gdb) run < input.txt
Starting program: /home/ken/Desktop/dynamic prog/maxsub < input.txt
Breakpoint 1, main () at maxsub.c:14
14      scanf("%d", &T);
(gdb) n
16      for (i = 0; i < T; i++) {                   // Each testcase
(gdb) n
17          scanf("%d", &N);                    // Get size of array
(gdb) n
18          int integers[N];                    // Allocate memory
(gdb) n
20          for (j = 0; j < N; j++) {               // Fill up array
(gdb) n
21              scanf("%d", &integers[j]);
(gdb) n
< SNIPPED PART > this is basically where for loop repeats N times
(gdb) n
20          for (j = 0; j < N; j++) {               // Fill up array
(gdb) n
24          printf("%d %d\n", bestSum1(N, integers), bestSum2(N,     integers));    // Print Answer
/** START TAKING A LOOK HERE, this is where the unusual thing happens: **/
(gdb) print integers[0]     <---- INITIAL VALUE
$1 = 2                     
(gdb) print integers        <---- ADDRESS OF ARRAY
$2 = 0x7fffffffdea0   
(gdb) n

Breakpoint 2, bestSum1 (size=6, integers=0x7fffffffdea0) at maxsub.c:33
33          best_sum = - INFINITY,                  // Best Sum Found
(gdb) print integers[0]     <---- VALUE CHANGED
$3 = -5
(gdb) print integers        <---- SAME ADDRESS
$4 = (int *) 0x7fffffffdea0
(gdb)

完整源代码：

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define INFINITY INT_MAX

int bestSum1(int size, int integers[size]);
int bestSum2(int size, int integers[size]);
int compare(const void* p1, const void* p2);

int main() {
    int T, N,   // testcases (T) and array size (N)
        i, j;   // Declare loop counters

    scanf("%d", &T);

    for (i = 0; i < T; i++) {               // Each testcase
        scanf("%d", &N);                    // Get size of array
        int integers[N];                    // Allocate memory

        for (j = 0; j < N; j++) {           // Fill up array
            scanf("%d", &integers[j]);
        }

        // bestSum1 gets strictly contiguous subsequence with largest sum
        // bestSum2 gets subset of numbers with largest sum
        printf("%d %d\n", bestSum1(N, integers), bestSum2(N, integers));    
    }

    return 0;
}

int bestSum1(int size, int integers[size]) {
    // CONTIGUOUS SUBSEQUENCE WITH LARGEST SUM
    int curr_sum = 0,                   // Current Sum
        best_sum = - INFINITY,          // Best Sum Found
        value,                          // Where we store value
        i;                              // Loop Counter

    for (i = 0; i < size; i++) {
            value = curr_sum + integers[i];
        if (value <= 0) {
            curr_sum = 0;
        } else if (value > best_sum) {
            best_sum = value;
            curr_sum = best_sum;
        }
    }

    return best_sum;
}

int bestSum2(int size, int integers[size]) {
    // NON-CONTIGUOUS SUBSEQUENCE WITH LARGEST SUM
    qsort(integers, size, sizeof(int), compare);

    int sum = -1,
        i;

    for (i = 0; i < size; i++) {
        if (integers[i] > 0) {
            sum += integers[i];
        }
    }

    return sum == -1 ? integers[size - 1] : sum;
}

int compare(const void* p1, const void* p2) {
    return *((int*) p1) - *((int*) p2);
}

所以我想知道如何解决这个问题以及为什么会这样。

传递给函数时整数数组更改内容

0 个答案: