我无法计算出两个maxsubarray函数的运行时间。 （位于代码底部） 它给我的输出：
Inputsize: 101 Time using Brute Force:0 Time Using DivandCon: 12
第二次使用clock（）是正确的但是对于第一个差异diff1它只是给了我0而我不确定为什么？
修改：修改代码。
编辑2：添加了输出。
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <limits.h>
using namespace std;
int Kedane(int a[], int size)
{
int max_so_far = 0, max_ending_here = 0;
int i;
for(i = 0; i < size; i++)
{
max_ending_here = max_ending_here + a[i];
if(max_ending_here < 0)
max_ending_here = 0;
if(max_so_far < max_ending_here)
max_so_far = max_ending_here;
}
return max_so_far;
}
int BruteForce(int array[],int n)
{
int sum,ret=0;
for(int j=-1;j<=n-2;j++)
{
sum=0;
for(int k=j+1;k<+n-1;k++)
{
sum+=array[k];
if(sum>ret)
{
ret=sum;
}
}
}
return ret;
}
//------------------------------------------------------
// FUNCTION WHICH FINDS MAX OF 2 INTS
int max(int a, int b) { return (a > b)? a : b; }
// FUNCTION WHICH FINDS MAX OF 3 NUMBERS
// CALL MAX FUNCT FOR 2 VARIS TWICE!
int max(int a, int b, int c) { return max(max(a, b), c); }
// WORKS OUT FROM MIDDLE+1->RIGHT THE MAX SUM &
// THE MAX SUM FROM MIDDLE->LEFT + RETURNS SUM OF THESE
int maxCrossingSum(int arr[], int l, int m, int h)
{
int sum = 0; // LEFT OF MID
int LEFTsum = INT_MIN; // INITIALLISES SUM TO LOWEST POSSIBLE INT
for (int i = m; i >= l; i--)
{
sum = sum + arr[i];
if (sum > LEFTsum)
LEFTsum = sum;
}
sum = 0; // RIGHT OF MID
int RIGHTsum = INT_MIN;
for (int i = m+1; i <= h; i++)
{
sum = sum + arr[i];
if (sum > RIGHTsum)
RIGHTsum = sum;
}
// RETURN SUM OF BOTH LEFT AND RIGHT SIDE MAX'S
return LEFTsum + RIGHTsum;
}
// Returns sum of maxium sum subarray in aa[l..h]
int maxSubArraySum(int arr[], int l, int h)
{
// Base Case: Only one element
if (l == h)
return arr[l];
// Find middle point
int m = (l + h)/2;
/* Return maximum of following three possible cases
a) Maximum subarray sum in left half
b) Maximum subarray sum in right half
c) Maximum subarray sum such that the subarray crosses the midpoint */
return max(maxSubArraySum(arr, l, m),
maxSubArraySum(arr, m+1, h),
maxCrossingSum(arr, l, m, h));
}
// DRIVER
int main(void)
{
std::srand (time(NULL));
// CODE TO FILL ARRAY WITH RANDOMS [-50;50]
int size=30000;
int array[size];
for(int i=0;i<=size;i++)
{
array[i]=(std::rand() % 100) -50;
}
// TIMING VARI'S
clock_t t1,t2;
clock_t A,B;
clock_t K1,K2;
volatile int mb, md, qq;
//VARYING ELEMENTS IN THE ARRAY
for(int n=101;n<size;n=n+100)
{
t1=clock();
mb=BruteForce(array,n);
t2=clock();
A=clock();
md=maxSubArraySum(array, 0, n-1) ;
B=clock();
K1=clock();
qq=Kedane(array, n);
K2=clock();
cout<< n << "," << (double)t2-(double)t1 << ","<<(double)B-(double)A << ","<<(double)K2-(double)K1<<endl;
}
return 0;
}
101,0,0,0
201,0,0,0
301,1,0,0
401,0,0,0
501,0,0,0
601,0,0,0
701,0,0,0
801,1,0,0
901,1,0,0
1001,0,0,0
1101,1,0,0
1201,1,0,0
1301,0,0,0
1401,1,0,0
1501,1,0,0
1601,2,0,0
1701,1,0,0
1801,2,0,0
1901,1,1,0
2001,1,0,0
2101,2,0,0
2201,3,0,0
2301,2,0,0
2401,3,0,0
2501,3,0,0
2601,3,0,0
2701,4,0,0
2801,4,0,0
2901,4,0,0
3001,4,0,0
3101,4,0,0
3201,5,0,0
3301,5,0,0
3401,6,0,0
3501,5,0,0
3601,6,0,0
3701,6,0,0
3801,8,0,0
3901,7,0,0
4001,8,0,0
4101,7,0,0
4201,10,1,0
4301,9,0,0
4401,8,0,0
4501,9,0,0
4601,10,0,0
4701,11,0,0
4801,11,0,0
4901,11,0,0
5001,12,0,1
5101,11,1,0
5201,13,0,0
5301,13,0,0
5401,15,0,0
5501,14,0,0
5601,16,0,0
5701,15,0,0
5801,15,1,0
5901,16,0,0
6001,17,0,0
6101,18,0,0
6201,18,0,0
6301,19,0,0
6401,21,0,0
6501,19,0,0
6601,21,1,0
6701,20,0,0
6801,22,0,0
6901,23,0,0
7001,22,0,0
7101,24,0,0
7201,26,0,0
7301,26,0,0
7401,24,1,0
7501,26,0,0
7601,27,0,0
7701,28,0,0
7801,28,0,0
7901,30,0,0
8001,29,0,0
8101,31,0,0
8201,31,1,0
8301,35,0,0
8401,33,0,0
8501,35,0,0
8601,35,1,0
8701,35,0,0
8801,36,1,0
8901,37,0,0
9001,38,0,0
9101,39,0,0
9201,41,1,0
9301,40,0,0
9401,41,0,0
9501,42,0,0
9601,45,0,0
9701,45,0,0
9801,44,0,0
9901,47,0,0
10001,47,0,0
10101,48,0,0
10201,50,0,0
10301,51,0,0
10401,50,0,0
10501,51,0,0
10601,53,0,0
10701,55,0,0
10801,54,0,0
10901,56,0,0
11001,57,0,0
11101,56,0,0
11201,60,0,0
11301,60,0,0
11401,61,1,0
11501,61,1,0
11601,63,0,0
11701,62,1,0
11801,66,1,0
11901,65,0,0
12001,68,1,0
12101,68,0,0
12201,70,0,0
12301,71,0,0
12401,72,0,0
12501,73,1,0
12601,73,1,0
12701,76,0,0
12801,77,0,0
12901,78,1,0
13001,79,1,0
13101,80,0,0
13201,83,0,0
13301,82,0,0
13401,86,0,0
13501,85,1,0
13601,86,0,0
13701,89,0,0
13801,90,0,1
13901,90,0,0
14001,91,0,0
14101,97,0,0
14201,93,0,0
14301,96,0,0
14401,99,0,0
14501,100,0,0
14601,101,0,0
14701,101,0,0
14801,103,1,0
14901,104,0,0
15001,107,0,0
15101,108,0,0
15201,109,0,0
15301,109,0,0
15401,114,0,0
15501,114,0,0
15601,115,0,0
15701,116,0,0
15801,119,0,0
15901,118,0,0
16001,124,0,0
16101,123,1,0
16201,123,1,0
16301,125,0,0
16401,127,1,0
16501,128,1,0
16601,131,0,0
16701,132,0,0
16801,134,0,0
16901,134,1,0
17001,135,1,0
17101,139,0,0
17201,139,0,0
17301,140,1,0
17401,143,0,0
17501,145,0,0
17601,147,0,0
17701,147,0,0
17801,150,1,0
17901,152,1,0
18001,153,0,0
18101,155,0,0
18201,157,0,0
18301,157,1,0
18401,160,0,0
18501,160,1,0
18601,163,1,0
18701,165,0,0
18801,169,0,0
18901,171,0,1
19001,170,1,0
19101,173,1,0
19201,178,0,0
19301,175,1,0
19401,176,1,0
19501,180,0,0
19601,180,1,0
19701,182,1,0
19801,184,0,0
19901,187,1,0
20001,188,1,0
20101,191,0,0
20201,192,1,0
20301,193,1,0
20401,195,0,0
20501,199,0,0
20601,200,0,0
20701,201,0,0
20801,209,1,0
20901,210,0,0
21001,206,0,0
21101,210,0,0
21201,210,0,0
21301,213,0,0
21401,215,1,0
21501,217,1,0
21601,218,1,0
21701,221,1,0
21801,222,1,0
21901,226,1,0
22001,225,1,0
22101,229,0,0
22201,232,0,0
22301,233,1,0
22401,234,1,0
22501,237,1,0
22601,238,0,1
22701,243,0,0
22801,242,1,0
22901,246,1,0
23001,246,0,0
23101,250,1,0
23201,250,1,0
23301,254,1,0
23401,254,0,0
23501,259,0,1
23601,260,1,0
23701,263,1,0
23801,268,0,0
23901,266,1,0
24001,271,0,0
24101,272,1,0
24201,274,1,0
24301,280,0,1
24401,279,0,0
24501,281,0,0
24601,285,0,0
24701,288,0,0
24801,289,0,0
24901,293,0,0
25001,295,1,0
25101,299,1,0
25201,299,1,0
25301,302,0,0
25401,305,1,0
25501,307,0,0
25601,310,1,0
25701,315,0,0
25801,312,1,0
25901,315,0,0
26001,320,1,0
26101,320,0,0
26201,322,0,0
26301,327,1,0
26401,329,0,0
26501,332,1,0
26601,339,1,0
26701,334,1,0
26801,337,0,0
26901,340,0,0
27001,341,1,0
27101,342,1,0
27201,347,0,0
27301,348,1,0
27401,351,1,0
27501,353,0,0
27601,356,1,0
27701,360,0,1
27801,361,1,0
27901,362,1,0
28001,366,1,0
28101,370,0,1
28201,372,0,0
28301,375,1,0
28401,377,1,0
28501,380,0,0
28601,384,1,0
28701,384,0,0
28801,388,1,0
28901,391,1,0
29001,392,1,0
29101,399,1,0
29201,399,0,0
29301,404,1,0
29401,405,0,0
29501,409,1,0
29601,412,2,0
29701,412,1,0
29801,422,1,0
29901,419,1,0
答案 0 :(得分：0)
从不使用BruteForce和maxSubArraySum的返回值，这为编译器提供了很多优化。
在我的机器上，例如，使用clang -O3将对BruteForce的调用减少为矢量副本，而不是其他内容。
强制评估这些函数的一种方法是将结果写入volatile变量：
volatile int mb, md;
// ...
mb = BruteForce(array, n);
// ...
md = maxSubArraySum(array, 0, n-1);
由于变量是易变的，所以必须存储赋值右侧给出的值，尽管没有任何其他副作用，这会阻止编译器优化计算。