Moo Volume
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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.
* 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).
There are five cows at locations 1, 5, 3, 2, and 4.
515324
40
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.
代码:
#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define pi acos(-1.0)#define inf 0x3f3f3f#define M 50005int n;int a[10002];int main(){ int i,j; ll ans=0; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%d",&a[i]); } sort(a+1,a+1+n); for(i=1;i<=n;i++) { for(j=1;j<=i;j++) ans+=a[i]-a[j]; for(j=i+1;j<=n;j++) ans+=a[j]-a[i]; } printf("%I64d",ans); return 0;}
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