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. GCD | light_ink_dots# | TL | 1ms | 3868kb | C++23 | 643b | 2024-12-14 19:16:04 | 2024-12-14 19:16:06 |
Judging History
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