【贪心】习题集

122. 糖果传递

写法一：

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;

const int N = 1e6 + 10;

int n;
LL s[N], c[N];

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++ )
    {
        scanf("%lld", &s[i]);
        s[i] += s[i - 1];
    }

    LL b = s[n] / n;
    int k = 0;
    for(int i = 1; i < n; i ++ ) c[k ++ ] = i * b - s[i];
    c[k ++ ] = 0;

    nth_element(c, c + k / 2, c + k);
    LL res = 0;
    for(int i = 0; i < k; i ++ )
        res += abs(c[i] - c[k / 2]);

    printf("%lld\n", res);
    return 0;
}

作者：Once.
链接：https://www.acwing.com/activity/content/code/content/3430736/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

写法二

#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

typedef long long LL;

const int N = 1e6 + 10; 

int n;
int a[N];
LL c[N];
LL sum;

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++ ) 
    {
        scanf("%d", &a[i]);
        sum += a[i];
    }    
    LL avg = sum / n;

    for(int i = n; i > 1; i -- )
    {
        c[i] = c[i + 1] + avg - a[i];
    }
    c[1] = 0;

    sort(c + 1, c + n + 1);

    LL res = 0;
    for(int i = 1; i <= n; i ++ ) res += abs(c[i] - c[(i + 1) / 2]);
    printf("%lld\n", res);
    return 0;
}

作者：Once.
链接：https://www.acwing.com/activity/content/code/content/1774442/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

写法三

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 1e6 + 10;

int n;
LL s[N], c[N];

int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++ )
    {
        scanf("%lld", &s[i]);
        s[i] += s[i - 1];
    }

    LL avg = s[n] / n;
    for(int i = 2; i <= n; i ++ ) c[i] = (i - 1) * avg - (s[i] - s[1]);

    sort(c + 1, c + n + 1);

    LL md = c[(n + 1) / 2];
    LL res = 0;
    for(int i = 1; i <= n; i ++ ) res += abs(c[i] - md);
    printf("%lld\n", res);
    return 0;
}

作者：Once.
链接：https://www.acwing.com/activity/content/code/content/3430855/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。