cf 1216d

https://codeforc.es/problemset/problem/1216/D 

贪心：

找出最大的a[i]，令为t，即假设这个t就是原来的x

然后b[i]=t-a[i];  b[i]表示每个缺了的数量。  

 

最后求一个所有b[i]的最大公约数。 

答案就是累加b[i]/gcd  

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int const N = 200000 + 10;
int a[N], b[N], n, ta;
int gcd(int a, int b)
{
    return b ? gcd(b, a % b) : a;
}
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        ta = max(ta, a[i]);
    }
    for (int i = 1; i <= n; i++)
        b[i] = ta - a[i];
    int g = b[1];
    for (int i = 2; i <= n; i++)
        g = gcd(g, b[i]);
    ll ans = 0;
    for (int i = 1; i <= n; i++)
        ans += b[i] / g;
    printf("%lld %d\n", ans, g);
    return 0;
}