[codeforces 1362A] Johnny and Ancient Computer 倍数关系用2的幂次统计

Codeforces Round #647 (Div. 2) - Thanks, Algo Muse!  参与排名人数12044

[codeforces 1362A]    Johnny and Ancient Computer   倍数关系用2的幂次统计

总目录详见https://blog.csdn.net/mrcrack/article/details/103564004

在线测评地址https://codeforces.com/contest/1362/problem/A

Problem	Lang	Verdict	Time	Memory
A - Johnny and Ancient Computer	GNU C++17	Accepted	31 ms	0 KB

样例分析如下

10 5
10=5×2^1
2的幂次是1
1<=3
输出1

11 44
44=11×2^2
2的幂次是2
2<=3
输出1

17 21
21无法用17乘2的幂次来表示
输出-1

1 1
1==1
输出0

96 3
96=3×2^5
2的幂次是5
5=3+2
输出2

2 128
128=2*2^6
2的幂次是6
6=3+3
输出2


1001 1100611139403776
1100611139403776=1001*2^40
2的幂次是40
40=3*13+1
输出14

1000000000000000000 1000000000000000000
1000000000000000000==1000000000000000000
输出0

7 1
7无法用1乘2的幂次来表示
输出-1

10 8
10无法用8乘2的幂次来表示
输出-1

AC代码如下

#include 
#define LL long long
int main(){
	int t,cnt,flag;
	LL a,b,c;
	scanf("%d",&t);
	while(t--){
		scanf("%lld%lld",&a,&b);
		if(a==b){printf("0\n");continue;}
		if(a