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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 |
---|---|---|---|---|---|---|---|---|---|
#738033 | #9622. 有限小数 | huo_hua_ya# | TL | 0ms | 3708kb | C++23 | 1.1kb | 2024-11-12 17:31:17 | 2024-11-12 17:31:17 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e6 + 7;
const int mod = 998244353;
//求解不定方程 ax+by=c 的最小整数解 x
ll exgcd(ll a, ll b, ll &x, ll &y)
{
if (!b) {x = 1; y = 0; return a;}
ll ans = exgcd(b, a % b, x, y);
ll tmp = x;
x = y;
y = tmp - a / b * y;
return ans;
}
ll x, y;
ll cal(ll a, ll b, ll c)
{
x = 0,y = 0;
ll ans = exgcd(a, b, x, y);
if (c % ans)return -1;
x *= c / ans; b /= ans;
// y *= c / ans;
if (b < 0)b = -b;
x = (x % b + b) % b;
return x;
}
void solve()
{
int a,b;
cin >> a >> b;
int p = b,k = 1;
while(p%2==0){
p/=2;
k*=2;
}
while(p%5==0){
p/=5;
k*=5;
}
ll c = cal(k,a,p);
ll t = (p - 1ll * k * c) / a;
int id = 2;
while(t < 0){
c = cal(k,a,p*id);
t = (p *id - 1ll *k *c)/a;
}
ll d = t *p;
cout << c << " " <<d << '\n';
}
int main()
{
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
cin >> _;
while (_--)
solve();
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 0ms
memory: 3708kb
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
Time Limit Exceeded
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...