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=100010;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7,STA=8000010;
//const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=1;
    LL sum,ans,rev,a;
    rev=MOD/2+1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        sum=0;
        for(i=1;i<=n;i++){
            scanf("%d",&num[i]);
            sum=(sum+num[i])%MOD;
        }
        sort(num+1,num+n+1);
        ans=0;num[0]=1;
        for(i=1;i<=n;i++){
            a=num[i]-num[i-1]+1;
            LL t=(a-2)*(n-i+1)%MOD;
            ans=(ans+(a-1)*(sum+sum-t)%MOD*rev%MOD*(n-i)%MOD)%MOD;
            sum=(sum-(a-1)*(n-i+1)%MOD-1)%MOD;
            ans=(ans+n-i)%MOD;
        }

        printf("Case %d: %lld
",ca++,(ans+MOD)%MOD);
    }
    return 0;
}