Binary search/binary search (this method is only used for sequential arrangement)
The algorithm is an efficient way to find numbers
假设n个数 次数=log2n 如找2的32次个数，只需要32次
Steps:
1. Set left= the base of the initial digit (0), right= the base of the last digit, mid=(left+right)/2
2. Constantly fold in half to compare with the number you want to find (k)
3. If a[mid]>k, then mid is obviously to the left of k, left=mid+1; if a[mid]<k, then mid is obviously to the right of k, right=mid-1;
repeat 1, 2, 3 steps, when a[mid]=k, find the number k, that is, output mid base
For example, a[10]={1,2,3,4,5,6,7,8,9,10} Find a number in the array, and output its base.
If you find the number 7, the output shows the base 6
left=0 ,right=9,mid=(left+right)/2=4, at this time a[mid]=5<7,left=mid+1=5
left=5,right=9,mid=(left+right) /2=7, at this time a[mid]=8>7,right=mid-1=7
left=5,right=7,mid=(left+right)/2=6, at this time a[mid]= 7=7, if you use the normal search method to find the output mid
, it will take 7 times to find it from the first position, which greatly reduces the efficiency
#define _CRT_SECURE_NO_WARNINGS 1
int main(){
int a[10] = {
1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
printf("请输入你想要的数字：");
int k;
scanf("%d",&k);//输入你想要的数字
int left = 0;
int right = 9;
while (left <= right){
//比较左右大小来得到想要数字的底数
int mid = (left + right) / 2;
if (a[mid] < k){
left = mid + 1;
}
else if (a[mid]>k){
right = mid - 1;
}
else{
printf("它的底数是：%d\n", mid);
break;
}
}
if (left > right){
printf("找不到奥!\n");
}
return 0;
}