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。。
 1 //STATUS:C++_AC_2090MS_1716KB
 2 #include <functional>
 3 #include <algorithm>
 4 #include <iostream>
 5 //#include <ext/rope>
 6 #include <fstream>
 7 #include <sstream>
 8 #include <iomanip>
 9 #include <numeric>
10 #include <cstring>
11 #include <cassert>
12 #include <cstdio>
13 #include <string>
14 #include <vector>
15 #include <bitset>
16 #include <queue>
17 #include <stack>
18 #include <cmath>
19 #include <ctime>
20 #include <list>
21 #include <set>
22 //#include <map>
23 using namespace std;
24 //#pragma comment(linker,"/STACK:102400000,102400000")
25 //using namespace __gnu_cxx;
26 //define
27 #define pii pair<int,int>
28 #define mem(a,b) memset(a,b,sizeof(a))
29 #define lson l,mid,rt<<1
30 #define rson mid+1,r,rt<<1|1
31 #define PI acos(-1.0)
32 //typedef
33 typedef long long LL;
34 typedef unsigned long long ULL;
35 //const
36 const int N=100010;
37 const int INF=0x3f3f3f3f;
38 const int MOD=1e9+7,STA=8000010;
39 //const LL LNF=1LL<<60;
40 const double EPS=1e-8;
41 const double OO=1e15;
42 const int dx[4]={-1,0,1,0};
43 const int dy[4]={0,1,0,-1};
44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
45 //Daily Use ...
46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
56 //End
57 
58 int num[N];
59 int T,n;
60 
61 int main(){
62  //   freopen("in.txt","r",stdin);
63     int i,j,k,ca=1;
64     LL sum,ans,rev,a;
65     rev=MOD/2+1;
66     scanf("%d",&T);
67     while(T--)
68     {
69         scanf("%d",&n);
70         sum=0;
71         for(i=1;i<=n;i++){
72             scanf("%d",&num[i]);
73             sum=(sum+num[i])%MOD;
74         }
75         sort(num+1,num+n+1);
76         ans=0;num[0]=1;
77         for(i=1;i<=n;i++){
78             a=num[i]-num[i-1]+1;
79             LL t=(a-2)*(n-i+1)%MOD;
80             ans=(ans+(a-1)*(sum+sum-t)%MOD*rev%MOD*(n-i)%MOD)%MOD;
81             sum=(sum-(a-1)*(n-i+1)%MOD-1)%MOD;
82             ans=(ans+n-i)%MOD;
83         }
84 
85         printf("Case %d: %lld
",ca++,(ans+MOD)%MOD);
86     }
87     return 0;
88 }