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 ;

 }