CF思维联系–CodeForces - 222 C Reducing Fractions(数学+有技巧的枚举）
CF思维联系–CodeForces - 222 C Reducing Fractions(数学+有技巧的枚举）
ACM思维题训练集合 To confuse the opponents, the Galactic Empire represents fractions in an unusual format. The fractions are represented as two sets of integers. The product of numbers from the first set gives the fraction numerator, the product of numbers from the second set gives the fraction denominator. However, it turned out that the programs that work with fractions in this representations aren’t complete, they lack supporting the operation of reducing fractions. Implement this operation and the Empire won’t forget you.
Input The first input line contains two space-separated integers n, m (1 ≤ n, m ≤ 105) that show how many numbers the first set (the numerator) and the second set (the denominator) contain, correspondingly.
The second line contains n space-separated integers: a1, a2, …, an (1 ≤ ai ≤ 107) — the numbers that are multiplied to produce the numerator.
The third line contains m space-separated integers: b1, b2, …, bm (1 ≤ bi ≤ 107) — the numbers that are multiplied to produce the denominator.
Output Print the answer to the problem in the form, similar to the form of the input data. The number of values in the sets you print nout, mout must satisfy the inequality 1 ≤ nout, mout ≤ 105, and the actual values in the sets aout, i and bout, i must satisfy the inequality 1 ≤ aout, i, bout, i ≤ 107.
Separate the values in the lines by spaces. The printed fraction must be reduced, that is, there mustn’t be such integer x (x > 1), that the numerator and the denominator of the printed fraction are divisible by x. If there are several matching answers, print any of them.
Examples Input 3 2 100 5 2 50 10 Output 2 3 2 1 1 1 1 Input 4 3 2 5 10 20 100 1 3 Output 1 1 20 3 Note In the first test sample the numerator equals 1000, the denominator equals 500. If we reduce fraction 1000/500 by the greatest common divisor of the numerator and the denominator (by 500), we obtain fraction 2/1.
In the second test sample the numerator equals 2000, the denominator equals 300. If we reduce fraction 2000/300 by the greatest common divisor of the numerator and the denominator (by 100), we obtain fraction 20/3.
日常WA一天 不看跑的数据，我都不知道自己怎么错的，老天爷。我的输出超出了限制100001不能超过100000，我觉得那时候，那些没有过的，一定是这个原因，出题人真是丧心病狂。 第一个代码是错的，第二个是修改了的，换了方式。
#include <bits/stdc++.h>
using namespace std;
template <typename t>
void read(t &x)
{
char ch = getchar();
x = 0;
int f = 1;
while (ch < '0' || ch > '9')
f = (ch == '-' ? -1 : f), ch = getchar();
while (ch >= '0' && ch <= '9')
x = x * 10 + ch - '0', ch = getchar();
x *f;
}
bitset<100000010> v;
int prime[6000001];
int m = 0;
void primes(int n)
{
for (int i = 2; i * i <= n; i++)
{
if (!v[i])
{
for (int j = i * i; j <= n; j += i)
v[j] = 1;
}
}
for (int i = 2; i <= n; i++)
if (!v[i])
prime[m++] = i;
}
vector<int> a[4];
unordered_map<int, int> c, d;
int main()
{
int n, m, maxi = 0;
read(n), read(m);
primes(10000005);
for (int i = 0; i < n; i++)
{
int tem;
read(tem);
maxi = max(maxi, tem);
c[tem]++;
}
for (int i = 0; i < m; i++)
{
int tem;
read(tem);
maxi = max(maxi, tem);
d[tem]++;
}
//cout << 1 << endl;
for (int i = 0; prime[i] <= maxi; i++)
{
// cout<<i<<endl;
int cnt = 0, ans = 0, cnt2 = 0;
int flag = 1;
for (auto po = c.begin(); po != c.end();)
{
// cout<<1<<endl;
pair<int, int> tem = *po;
cnt = 0;
if (tem.first < prime[i])
{
po++;
continue;
}
else
{
flag = 0;
while (tem.first % prime[i] == 0)
{
tem.first /= prime[i];
cnt++;
//cout<<i<<endl;
}
cnt *= tem.second;
auto pi = po;
po++;
c.erase(pi);
if (tem.first != 1)
c[tem.first] += tem.second;
}
ans += cnt;
}
cnt2 = ans;
ans = 0;
for (auto po = d.begin(); po != d.end();)
{
pair<int, int> tem = *po;
cnt = 0;
if (tem.first < prime[i])
{
po++;
continue;
}
else
{
flag = 0;
while (tem.first % prime[i] == 0)
{
tem.first /= prime[i];
cnt++;
//cout<<i<<endl;
}
cnt *= tem.second;
auto pi = po;
po++;
d.erase(pi);
if (tem.first != 1)
d[tem.first] += tem.second;
}
ans += cnt;
}
cnt = cnt2 - ans;
if (cnt == 0)
continue;
else if (cnt < 0)
{
cnt = -cnt;
int temp = 1;
int j = 0;
for (; j < cnt; j++)
{
temp *= prime[i];
if (temp * prime[i] > 10000000)
{
a[3].push_back(temp);
// cout << 1 << endl;
temp = 1;
}
}
a[3].push_back(temp);
}
else
{
int temp = 1;
int j = 0;
for (; j < cnt; j++)
{
temp *= prime[i];
if (temp * prime[i] > 10000000)
{
a[2].push_back(temp);
// cout << 1 << endl;
temp = 1;
}
}
a[2].push_back(temp);
}
if (flag)
break;
}
if (a[2].size() == 0)
a[2].push_back(1);
if (a[3].size() == 0)
a[3].push_back(1);
cout << a[2].size() << " " << a[3].size() << endl;
for (int i = 0; i < a[2].size(); ++i)
printf("%d ", a[2][i]);
puts("");
for (int i = 0; i < a[3].size(); ++i)
printf("%d ", a[3][i]);
puts("");
}#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int n, m, tot, a[100005], b[100005], z[10000005], pos[10000005], q[1000005], t1[1000005], t2[1000005];
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
for (int i = 1; i <= m; ++i)
scanf("%d", &b[i]);
for (int i = 2; i <= 10000000; ++i)
if (!z[i])
{
for (int j = i; j <= 10000000; j += i)
z[j] = i;
q[++tot] = i;
pos[i] = tot;
}
for (int i = 1; i <= n; ++i)
{
int k = a[i];
while (k != 1)
{
t1[pos[z[k]]]++;
k /= z[k];
}
}
for (int i = 1; i <= m; ++i)
{
int k = b[i];
while (k != 1)
{
t2[pos[z[k]]]++;
k /= z[k];
}
}
for (int i = 1; i <= tot; ++i)
{
t1[i] = min(t1[i], t2[i]);
t2[i] = t1[i];
}
printf("%d %d\n", n, m);
for (int i = 1; i <= n; ++i)
{
int k = a[i], p = a[i];
while (k != 1)
{
if (t1[pos[z[k]]])
{
p /= z[k];
t1[pos[z[k]]]--;
}
k /= z[k];
}
printf("%d ", p);
}
printf("\n");
for (int i = 1; i <= m; ++i)
{
int k = b[i], p = b[i];
while (k != 1)
{
if (t2[pos[z[k]]])
{
p /= z[k];
t2[pos[z[k]]]--;
}
k /= z[k];
}
printf("%d ", p);
}
printf("\n");
return 0;
}腾讯云开发者
