1246. 等差数列
等差数列
思路
样例输入:
5
2 6 4 10 20
样例输出:
10
代码O(nlogn):
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N = 100000;
int n;
int a[N];
//欧几里得求最大公约数
int gcd(int a, int b)
{
return b ? gcd(b, a % b) : a;
}
int main()
{
cin >> n;
for(int i = 1; i <= n; i ++ )
cin >> a[i];
// A1∼AN 并不一定是按等差数列中的顺序给出 -> 排序
sort(a + 1, a + n + 1);
//求公差的最大公约数开始check
int d = 0;
int t = a[n] - a[1];
for(int i = 1; i < n; i ++ )
d = gcd(d, a[i + 1] - a[i]);
int res;
//如果公差为0 即该等差数列的最小个数为n
if(!d) res = n;
else res = t / d + 1;
cout << res << endl;
return 0;
}
1223. 最大比例
最大比例
思路:
样例输入:
3
549755813888 524288 2
样例输出:
4/1
代码:
#include<iostream>
#include<algorithm>
using namespace std;
const int N = 105;
typedef long long LL;
LL x[N], up[N], down[N];
LL gcd(LL a, LL b)
{
return b ? gcd(b, a % b) : a;
}
//辗转相减法
LL getsub(LL a, LL b)
{
if(a < b) swap(a, b);
if(b == 1) return a;
return getsub(b, a / b);
}
int main()
{
int n;
cin >> n;
for(int i = 0; i < n; i ++ ) cin >> x[i];
sort(x, x + n);
//去重
LL m = 0;
for(int i = 0; i < n; i ++ )
{
LL d = gcd(x[0], x[i]);
if(x[i] == x[i + 1]) continue;
//约分
up[m] = x[i] / d;
down[m ++ ] = x[0] / d;
}
LL a = up[0], b = down[0];
for(int i = 1; i < m; i ++ )
{
a = getsub(a, up[i]);
b = getsub(b, down[i]);
}
cout << a << "/" << b << endl;
return 0;
}