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\n",ca++,(ans+MOD)%MOD);

86     }

87     return 0;

88 }