[BZOJ] 1679: [Usaco2005 Jan]Moo Volume 牛的呼声
1679: [Usaco2005 Jan]Moo Volume 牛的呼声
Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 1101 Solved: 573
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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.
Input
* Line 1: N * Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).
Output
* Line 1: A single integer, the total volume of all the MOOs.
Sample Input
1
5
3
2
4
INPUT DETAILS:
There are five cows at locations 1, 5, 3, 2, and 4.
Sample Output
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.
HINT
Source
Analysis
水水水题??
不是很清楚怎么优化,只把空间优化到线性
反正随便就过了qwq
递推什么的不会啊qwq
Code
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 using namespace std; 5 6 long long pos[100000],cnt[100000],tot,n; 7 8 int main(){ 9 scanf("%lld",&n); 10 for(int i = 1;i <= n;i++) scanf("%lld",&pos[i]); 11 sort(pos+1,pos+1+n); 12 13 for(int i = 1;i <= n-1;i++){ 14 for(int j = i;j <= n-1;j++){ 15 cnt[j] += n-j; 16 }tot += cnt[i]*(pos[i+1]-pos[i]); 17 } 18 19 printf("%lld",tot*2); 20 21 return 0; 22 }