练习1:求1!+2!+...+10!
一般算法:双层循环,外层1~10,内层计算每个数的阶乘,在外层把阶乘相加。
int main() { int i = 0; int j = 0; int jc = 1; int sum = 0; for (i = 1; i <= 10; i++) { jc = 1;// for (j = 1; j <= i; j++) { jc = jc * j; } sum = sum + jc; } printf("%d %d\n", jc,sum); return 0; }
算法优化:1~10递增,其阶乘比上一个多乘一个,在这过程中相加,一层循环即可。
int main() { int i = 0; int jc = 1; int sum = 0; for (i = 1; i <= 10; i++) { jc = jc * i; sum = sum + jc; } printf("%d %d\n", jc, sum); return 0; }
练习2:有序数组查找, 报数,在数组中查找,输出其下标
一般算法:遍历数组,if条件判断输出。
int main()
{
int input = 0;
scanf("%d", &input);
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 };
int len = sizeof(arr) / sizeof(arr[0]); //求出数组长度
printf("%d\n", sizeof(arr));
printf("len=%d\n", len);
int i = 0;
for (i = 0; i < len; i++)
{
if (input == arr[i])
{
printf("下标为:%d\n", i);
break;
}
}
if (i == len)
{
printf("i=%d\n",i);
printf("out of this area");
}
return 0;
}
未能利用 “有序” 这个条件
算法优化:一层循环用 二分法 折半查找,每次查找少一半数据,效率高。
int main()
{
int input = 0;
scanf("%d", &input);
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 };
int len = sizeof(arr) / sizeof(arr[0]); //求出数组长度
printf("%d\n", sizeof(arr));
printf("len=%d\n", len);
int left = 0;
int right = len - 1;
int mid = 0;
while (left <= right)
{
//mid = (left + right) / 2;
mid = left + (right - left) / 2; //此写法避免left和right数值过大 求平均值出错
if (arr[mid] < input)
{
left = mid + 1;
}
if (arr[mid] > input)
{
right = mid - 1;
}
if (input == arr[mid])
{
printf("找到了,下标为:%d\n", mid);
break;
}
}
if (left > right)
printf("out of this area");
return 0;
}