QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#814658#9868. GCDlight_ink_dots#TL 1ms3868kbC++23643b2024-12-14 19:16:042024-12-14 19:16:06

Judging History

你现在查看的是最新测评结果

  • [2024-12-14 19:16:06]
  • 评测
  • 测评结果:TL
  • 用时:1ms
  • 内存:3868kb
  • [2024-12-14 19:16:04]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
int T,a,ans,t;
long long b;
int bgcd(int a,int b){
	int at=__builtin_ctz(a),bt=__builtin_ctz(b),t=min(at,bt);
	b>>=bt;
	while(a){
		a>>=at;
		int c=a-b;
		b=min(a,b),a=abs(c),at=__builtin_ctz(a);
	}
	return b<<t;
}
void dfs(int a,int b,int cnt){
	if(cnt>=ans)
		return ;
	if(a==0||b==0){
		ans=min(ans,cnt+1);
		return ;
	}
	int g=bgcd(a,b);
	dfs(a,b-g,cnt+1);
	dfs(a-g,b,cnt+1);
}
int main(){
	scanf("%d",&T);
	while(T--){
		scanf("%d%lld",&a,&b),t=0;
		for(int i=0;i<=20;i++)
			if(a>=(1<<i))
				t+=2;
		ans=t,dfs(a,b,0);
		printf("%d\n",ans);
	}
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3868kb

input:

3
3 4
12 20
114 514

output:

3
4
6

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 1ms
memory: 3796kb

input:

990
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
1 11
1 12
1 13
1 14
1 15
1 16
1 17
1 18
1 19
1 20
1 21
1 22
1 23
1 24
1 25
1 26
1 27
1 28
1 29
1 30
1 31
1 32
1 33
1 34
1 35
1 36
1 37
1 38
1 39
1 40
1 41
1 42
1 43
1 44
1 45
1 46
1 47
1 48
1 49
1 50
1 51
1 52
1 53
1 54
1 55
1 56
1 57
1 58
1 59
1 60
2 3
2 4
2...

output:

2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
2
3
4
2
3
3
2
3
4
2
3
3
2
3
4
2
3
3
2
3
4
2
3
3
2
3
4
2
3
3
2
3
4
2
...

result:

ok 990 lines

Test #3:

score: -100
Time Limit Exceeded

input:

2
4859 299556476011016293
4859 911621905353047038

output:


result: