文章地址:http://henuly.top/?p=707
Description:
Give two positive integer c, n. You need to find a pair of integer (a,b) satisfy 1<=a,b<=n and the greatest common division of a and b is c.And you need to maximize the product of a and b
Input:
The first line has two positive integer c,n
Output:
Output the maximum product of a and b.
If there are no such a and b, just output -1
Sample Input:
2 4
Sample Output:
8
Hint:
1<=c,n<=10^9
题目链接
,求 。
, 和 互质,找到最大的 即可。
,而 一定与 互质,
注意特判 以及无解的情况
AC代码:
#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
#define XDebug(x) cout << #x << "=" << x << endl;
#define ArrayDebug(x,i) cout << #x << "[" << i << "]=" << x[i] << endl;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e7 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
template <class T>
inline bool read(T &ret) {
char c;
int sgn;
if (c = getchar(), c == EOF) {
return 0;
}
while (c != '-' && (c < '0' || c > '9')) {
c = getchar();
}
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9') {
ret = ret * 10 + (c - '0');
}
ret *= sgn;
return 1;
}
template <class T>
inline void out(T x) {
if (x > 9) {
out(x / 10);
}
putchar(x % 10 + '0');
}
int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
#endif
ll c, n;
read(c); read(n);
if (c > n) {
printf("-1\n");
}
else if (n / c == 1) {
printf("%lld\n", c * c);
}
else {
printf("%lld\n", c * c * (n / c) * (n / c - 1));
}
#ifndef ONLINE_JUDGE
fclose(stdin);
fclose(stdout);
system("gedit out.txt");
#endif
return 0;
}