Day 7_1 binary search

Common binary search code is omitted ......
sometimes the next left and upper bounds right, calculating mid, left + right type int would exceed the scope of use
mid = left + (right - left)/2 Instead, to avoid overflow

Calculating approximation root 2, f (x) = x ^ 2; x is incremented between 1-2, using the left and right closed closed interval

1. No. 2 approximation Root

#include<stdio.h>
const double eps = 1e-5;
double f(double x){
	return x*x;
}
double calsqrt(){
	double left = 1,right = 2,mid;
	while((right-left)>eps){
		mid = (left+right)/2;
		if(f(mid)>2){
			right = mid;
		} else{
			left = mid;
		}
	}
	return mid;
}
int main(){
	double k = calsqrt();
	printf("%.5f\n",k);
	return 0;
}

2. Fast power

Given three integers, A, B, m
A <10 ^. 9
B <10 ^ 18 is
. 1 <m <10 ^. 9
seeking a ^ b% m

1) written recursively

typedef long long LL;
LL binaryPow(LL a, LL b,LL m){
	if(b==0)	return 1;
	if(b%2==1)	return a * binaryPow(a,b-1,m);
	else{
		LL mul = binaryPow(a,b/2,m);
		return mul * mul % m;
	} 
}

2) the wording of Iteration

b writing binary conversion, e.g., 13 = 8 + 4 + 3 + 1 = 2 ^ 2 ^ 2 + 2 ^ 0 = 1101 (binary)
Shilling ans = 1, kanb binary bits whether the end of a, ans as a = ans * a;
then make a = a ^ 2, and b the right that, look as if a tail bit, is a continuing ans = ans * a
loop execution of the step (b just greater than 0)

LL binaryPow1(LL a, LL b,LL m){
	LL ans = 1;
	while(b>0){
		if(b & 1)
			ans = ans * a % m;
		a = a * a % m;
		b >>= 1; 
	}
	return ans;
}