poj 2231 Moo Volume
Moo Volume
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 15037 | Accepted: 4342 |
Description
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.
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int cows[10000];
int n;
int i,j;
__int64 sum=0;
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&cows[i]);
for(i=0;i<n;i++)
{
for(j=i+1;j<n;j++)
{
sum+=2*abs(cows[j]-cows[i]);
}
}
printf("%I64d\n",sum);
return 0;
}
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