HDU-4655 Cut Pieces 数学,贪心

时间：2022-04-22 21:50:26

　　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(ai)，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(ai)-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 ;

 }

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