首页 > 代码库 > BZOJ1679: [Usaco2005 Jan]Moo Volume 牛的呼声
BZOJ1679: [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: 723 Solved: 346
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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
Silver
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #define inf 100000000012 #define maxn 10000+10013 #define maxm 100000+10014 #define ll long long15 using namespace std;16 inline ll read()17 {18 ll x=0,f=1;char ch=getchar();19 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}20 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}21 return x*f;22 }23 ll a[maxn];24 int main()25 {26 freopen("input.txt","r",stdin);27 freopen("output.txt","w",stdout);28 ll n=read(),ans=0,sum;29 for(int i=1;i<=n;i++)a[i]=read();30 sort(a+1,a+n+1);sum=a[1];31 for(int i=2;i<=n;i++)32 {33 ans+=(a[i]*(i-1))-sum;sum+=a[i]; 34 }35 printf("%lld\n",2*ans);36 return 0;37 }