BNUOJ-29365 Join in tasks 简单数学

　　题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=29365

　　首先排序，然后维护一个后缀，等差求下和就可以了。。

 //STATUS:C++_AC_2090MS_1716KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 //#include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef long long LL;
 typedef unsigned long long ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=1e9+,STA=;
 //const LL LNF=1LL<<60;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 int num[N];
 int T,n;

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,k,ca=;
     LL sum,ans,rev,a;
     rev=MOD/+;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         sum=;
         for(i=;i<=n;i++){
             scanf("%d",&num[i]);
             sum=(sum+num[i])%MOD;
         }
         sort(num+,num+n+);
         ans=;num[]=;
         for(i=;i<=n;i++){
             a=num[i]-num[i-]+;
             LL t=(a-)*(n-i+)%MOD;
             ans=(ans+(a-)*(sum+sum-t)%MOD*rev%MOD*(n-i)%MOD)%MOD;
             sum=(sum-(a-)*(n-i+)%MOD-)%MOD;
             ans=(ans+n-i)%MOD;
         }

         printf("Case %d: %lld\n",ca++,(ans+MOD)%MOD);
     }
     return ;
 }