15- 1 << k 时的益出

扩展GCD-时间复杂性

题目：

计算循环语句的执行频次 for (i = A; i != B; i += C) x += 1;
其中A, B, C, i都是k位无符号整数。

输入：

A B C k, 其中0<k<32

输出：

输出执行频次数，如果是无穷，则输出“forever”

#include <iostream>
#include <algorithm> 
#include <cmath>
using namespace std;
//#define LL long long
typedef long long LL;
LL gcd(LL a, LL b)
{
	return b == 0 ? a : gcd(b, a%b);
}
 
LL ex_gcd(LL a, LL b, LL &x, LL &y)
{
	if (b == 0)
	{
		x = 1;
		y = 0;
		return a;
	}
	LL ans = ex_gcd(b, a%b, x, y);
	LL temp = x;
	x = y;
	y = temp - a / b*y;
	return ans;
}

int main(){
	LL A, B, C, k;
	cin >> A >> B >> C >> k;
	LL a = C,  n = B - A, x, y; //b = pow(2,k),
	//b改成b = 1 << k就会出错
	LL d = 1;
	LL b = d << k; 
	cout << "b: " << b << endl;
	int gc = gcd(a,b);
	if(A == 0 && B == 0){
		cout << 0 << endl;
		return 0;
	} 
	if(C == 0 || gc == 0 || n % gc != 0){
		cout << "forever" << endl;
		return 0;
	}
	ex_gcd(a,b,x,y);    //返回ax + by = gcd(a,b)的解 
	x = x * (n / gc);   //得到通解x即：ax + by = n 
	LL nn = b / gc;     //通解x的最小周期 
	x = (x % nn + nn) % nn; //得到最小解 
	cout << x << endl;
	return 0;
} 

　　应该是益出了：

#include <iostream>
#include <cmath>
using namespace std;
#define LL long long

int main()
{
	int k;
	cin >> k;
	LL a = 1;
	LL b = a << k;
	LL c = 1 << k;
	cout << "b: " << b << endl;
	cout << "c: " << c << endl;
	
	return 0;
}

/*
30
b: 1073741824
c: 1073741824

31
b: 2147483648
c: -2147483648

32
b: 4294967296
c: 1

33
b: 8589934592
c: 2

34
b: 17179869184
c: 4
*/