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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
#745850#9622. 有限小数Chen1098WA 11ms3920kbC++142.2kb2024-11-14 12:00:032024-11-14 12:00:04

Judging History

你现在查看的是最新测评结果

  • [2024-11-14 12:00:04]
  • 评测
  • 测评结果:WA
  • 用时:11ms
  • 内存:3920kb
  • [2024-11-14 12:00:03]
  • 提交

answer

#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,##__VA_ARGS__)
#define int long long
#define endl '\n'
#define pb push_back
#define pii pair<int,int>
using namespace std;
bool Men;
const int mod=998244353;
const int N=200005;

inline void write(int x){if(x<0) putchar('-'),x=-x;if(x>9) write(x/10);putchar(x%10+'0');return;}
inline int read() {int x=0,f=1; char ch=getchar();while(ch<'0'||ch>'9') { if(ch=='-') f=-1; ch=getchar();}while(ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;}
void add(int &x,int y){x=(x+y)%mod;}
int ksc(int m,int n,int p){int ans=0;while(n){if(n%2==1){n=n-1;ans=(ans+m)%p;}n=n/2;m=(m+m)%p;}return ans;}
int ksm(int base,int power,int p){int result=1;while(power>0){if(power&1){result=result*base%p;}power>>=1;base=(base*base)%p;}return result;}
int mod_inverse(int a){return ksm(a,mod-2,mod);}
int inv(int numerator,int denominator){int inverse_denominator = mod_inverse(denominator);int result=(1LL*numerator*inverse_denominator)%mod;return result;}
int a,b;
int g[1000000];
bool Mbe;
signed main(){
//	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
	int T=read();
	int cnt=0;
	for(int i=0;i<=100000;i++){
		for(int j=0;j<=100000;j++){
			if(pow(2,i)*pow(5,j)>1e9) break;
			g[++cnt]=pow(2,i)*pow(5,j);
		}
	}
	sort(g+1,g+1+cnt);
	while(T--){
		a=read(),b=read();
		int bb=b;
		while(bb%5==0){
			bb/=5;
		}
		while(bb%2==0){
			bb/=2;
		}
		if(bb==1){
			cout<<0<<' '<<1<<endl;
			continue;
		}
		
		
		int c=1e9,d=1e9;
		for(int q=1;q<=cnt;q++){
			int i=g[q];
			if(b*i>1e9) continue;
			int nowb=b*i,tot=1;
			while(nowb%2==0){
				if(nowb%2!=0) break;
				nowb/=2; 
				tot*=2;
			}
			while(nowb%5==0){
				if(nowb%5!=0) break;
				nowb/=5;
				tot*=5;
			}
			int nowa=a*i;
			nowb=b*i;
			int must=nowb/tot;
			for(int j=(int)((nowa+must-1)/must)-3;j<=((c+nowa)/must)+3;j++){
				int nowmust=must*j;
				if(nowmust<=nowa) continue;
//				cout<<nowmust-nowa<<endl;
				if(nowmust-nowa<c){
					c=nowmust-nowa;
					d=nowb;
				}
				else break;	
			}
			if(c==1) break;
//			cout<<c<<endl;
		}
		cout<<c<<' '<<d<<endl;
	}
/*

2 5 

2
28 53
2 35


1
1
*/
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3920kb

input:

4
1 2
2 3
3 7
19 79

output:

0 1
1 3
1 14
3 316

result:

ok 4 case(s)

Test #2:

score: -100
Wrong Answer
time: 11ms
memory: 3864kb

input:

10000
11 12
28 53
17 60
2 35
17 181
80 123
68 141
79 163
71 99
13 64
33 61
15 32
16 61
11 86
33 74
128 143
40 53
7 23
30 31
5 6
86 181
73 91
13 23
71 81
1 2
7 38
117 160
33 83
129 151
88 153
25 58
16 19
19 141
95 124
43 96
71 139
11 59
106 109
93 152
34 43
17 99
1 57
20 159
16 25
5 73
159 170
172 17...

output:

1 12
1 54272
1 60
1 350
1 231680000
23 3936
1 36096000
5 326
1 63360
0 1
1 31232
0 1
1 4880
1 10750
1 18500
1 11714560
1 331250
1 2944
1 31
1 6
1 289600000
1 455000
1 58880
1 51840
0 1
1 304
0 1
1 415
1 19328000
1 765000000
1 4640
1 608
1 72192
3 24800
1 192
3 347500
1 944
1 43600
1 1520
1 430000
1 ...

result:

wrong answer Jury found better answer than participant's 1 < 2 (Testcase 321)