本文实例分析了C语言连续子向量的最大和及时间度量，分享给大家供大家参考之用。具体方法如下：
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#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#define SCALE 3000
int maxnum(int a, int b);
int main(int argc, char const *argv[])
{
FILE *fp;
fp = fopen("maximum.in", "r");
// int x[] = {1,12,-11,10,-65,54,22,-9,21,5,48,5,-8,-2,56,54,-88,-5,2,-8,554,-56,35,-55,555,-65,-545,-23,48,-5,88,-56,16,-8};
int *x = (int *)malloc(sizeof(int)*(SCALE+1));
int xi = SCALE,a = 0,num_in = 0;
while(xi--){
fscanf(fp, "%d", &x[a++]);
}
clock_t start, end;
// ***Algorithm-1 cube***
start = clock();
int max = 0;
int length = SCALE;
int i,j,k;
for (i = 0; i < length; ++i)
{
for (j = i; j < length; ++j)
{
int sum = 0;
for (k = i; k <= j; ++k)
{
sum += x[k];
}
max = maxnum(max, sum);
}
}
// long num = 10000000L;
// while(num--);
end = clock();
double times = (double)(end - start)/CLOCKS_PER_SEC;
double dend = (double)end;
printf("\n***Algorithm-1 cube***\n");
printf("end: %f\n", dend);
printf("Time consuming: %f\n", times);
printf("%d\n", max);
// ***Algorithm-2 square***
start = clock();
max = 0;
for (i = 0; i < length; ++i)
{
int sum = 0;
for (j = i; j < length; ++j)
{
sum += x[j];
max = maxnum(max, sum);
}
}
end = clock();
times = (double)(end - start)/CLOCKS_PER_SEC;
dend = (double)end;
printf("\n***Algorithm-2 square***\n");
printf("end: %f\n", dend);
printf("Time consuming: %f\n", times);
printf("%d\n", max);
// ***Algorithm-3 linear***
start = clock();
max = 0;
int max_end_here = 0;
for (i = 0; i < length; ++i)
{
max_end_here = maxnum(max_end_here + x[i], 0);
max = maxnum(max, max_end_here);
}
end = clock();
times = (double)(end - start)/CLOCKS_PER_SEC;
dend = (double)end;
printf("\n***Algorithm-3 linear***\n");
printf("end: %f\n", dend);
printf("Time consuming: %f\n", times);
printf("%d\n", max);
free(x);
x = NULL;
return 0;
}
int maxnum(int a, int b)
{
return a > b ? a : b;
}
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感兴趣的朋友可以测试运行一下本文实例以加深理解。希望本文所述对大家C程序设计的学习有所帮助。