Moo Volume POJ - 2231
Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise.
FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.
FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.
Input
* Line 1: N
* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).
* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).
Output
There are five cows at locations 1, 5, 3, 2, and 4.
Sample Input
5 1 5 3 2 4
Sample Output
40
Hint
INPUT DETAILS:
There are five cows at locations 1, 5, 3, 2, and 4.
OUTPUT DETAILS:
Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.
There are five cows at locations 1, 5, 3, 2, and 4.
OUTPUT DETAILS:
Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.
求任意两个点之间的距离总和;
这题数据大肯定不能直接求,要找规律;
总共有(n-i-1)*(a[n-i-1]-a[i])算出了每两个点的距离
这总共是一半的距离和 所以要乘以2
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<algorithm> 5 #include<queue> 6 #include<cctype> 7 using namespace std; 8 long long a[10010]; 9 int main() { 10 int n; 11 while(scanf("%d",&n)!=EOF){ 12 for (int i=0 ;i<n ;i++) 13 scanf("%lld",&a[i]); 14 sort(a,a+n); 15 long long sum=0; 16 for (int i=0 ;i<n ;i++ ) 17 sum+=(n-1-i)*(a[n-1-i]-a[i]); 18 printf("%lld\n",sum*2); 19 } 20 return 0; 21 }