Moo Volume POJ - 2231

原题链接

考察:递推

思路:

        设f[i]是i与其他牛的噪音和,我们可以发现它与f[i+1]的关系是f[i+1] = f[i]+i*d-(n-i)*d(d为与i+1的距离).实际是指f[i+1]比f[i]多i个d的距离,而f[i]又比f[i+1]多计算了(n-i)个距离.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 typedef long long LL;
 7 const int N = 10010;
 8 LL f[N],a[N];
 9 int main()
10 {
11     int n;
12     scanf("%d",&n);
13     for(int i=1;i<=n;i++) scanf("%d",&a[i]);
14     sort(a+1,a+n+1);
15     for(int i=2;i<=n;i++) f[1]+=a[i]-a[1];
16     LL ans = f[1];
17     for(int i=1;i<n;i++)
18     {
19         LL d= a[i+1]-a[i];
20         f[i+1] = f[i]-(n-i)*d+i*d;
21         ans+=f[i+1];
22     }
23     printf("%lld\n",ans);
24     return 0;
25 }

标签：原题,int,LL,long,Volume,Moo,距离,POJ,include
来源： https://www.cnblogs.com/newblg/p/14428179.html