HDU-4655 Cut Pieces 数学,贪心

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4655

　　先不考虑相临的有影响，那么总数就是n*prod(a_i)，然后减去每个相邻的对总数的贡献Σ( Min(a[i],a[i+1])*prod(i-1)*prod(i+2) )，其中prod(i)为[1,i-1]这个区间a[i]的积，prod(i+2)表示的是[i+2,n]这个区间，答案就是n*prod(a_i)-Σ( Min(a[i],a[i+1])*prod(i-1)*prod(i+2) )，可以得到取得最大值是ai的排列就是最小，最大，次小，次大，例如:1 2 3 4 -> 1 4 2 3。然后就是一个DP转移了，方程很好写。。

 //STATUS:C++_AC_468MS_19068KB
 #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 __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD= ,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e30;
 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
 
 LL num[N],s[N];
 LL f[N],p[N];
 int T,n;
 
 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,mid;
     LL low,hig;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         for(i=;i<=n;i++)
             scanf("%I64d",&num[i]);
         sort(num+,num+n+);
         mid=(n+)>>;
         for(i=;i<=n;i+=){
             s[i]=num[(i+)>>];
             s[i+]=num[n-((i+)>>)+];
         }
         p[]=;
         for(i=;i<=n;i++)p[i]=(p[i-]*s[i])%MOD;
         f[]=s[];
         for(i=;i<=n;i++){
             low=Min(s[i-],s[i]);
             hig=Max(0LL,s[i]-s[i-]);
             f[i]=( low*(f[i-]+p[i-]-p[i-])%MOD+hig*(f[i-]+p[i-])%MOD )%MOD;
         }
 
         printf("%I64d\n",(f[n]+MOD)%MOD);
     }
     return ;
 }