Apple - Hdu5160

Problem Description

We are going to distribute apples to n children. Every child has his/her desired number of apple A1,A2,A3,⋯An ， Ai indicates the i-th child’s desired number of apple. Those children should stand in a line randomly before we distribute apples. Assume that the i-th position is occupied by pi−th child, i.e. from left to right the id of the children are p1,p2,p3,⋯pn ，So their desired number of apple are Ap1,Ap2,Ap3,⋯Apn . We will distribute Ap1 apples to the left most child，and for the i-th (i>1)child，if there exists a j which makes j < i and Apj>Api true, we will distribute no apples to him/her，otherwise we will distribute Api apples to him/ner. So for a certain permutation there exist a certain number of apples we should distribute. Now we want to know how many apples we should distribute for all possible permutation.
Note: two permutations are considered different if and only if there exists at least a position in which two children’s desired number of apple are different.

 

Input

Multi test cases，the first line contains an integer T which indicates the number of test cases. Then every case occupies two lines.
For each case, the first line contains an integer n which indicates there are n children.
The second line contain n integers A1,A2,A3,⋯An

indicate n children’s desired number of apple.
[Technique specification]
All numbers are integer
1<=T<=20
1<=n<=100000
1<=Ai<=1000000000

 

Output

For each case，output occupies one line，the output format is Case #x: ans, x is the data number starting from 1，ans is the total number of apple modular 1000000007.

 

Sample Input

2
2
2 3
3
1 1 2

 

Sample Output

Case #1: 8
Case #2: 9

Hint

For the second case, all possible permutation and

corresponding distributed apples are
(1,1,2),4
(1,2,1),3
(2,1,1),2
So the total number of apple is 2+3+4=9

 

Source

BestCoder Round #26

 

Recommend

heyang

 

简单题意

给你一堆数字，要计算所有不同排列的权值（两个排列为不一样的排列当且仅当至少有一个位置上的数字不同）

权值的计算方法：对于每个位置上的数字，要是这个数字之前不存在比这个数字大的数字，权值就加上这个数字

胡说题解

先排序，然后我们对于每一种数字，我们枚举有多少个被计算到权值里面，然后排列组合出这种情况的不同排列个数

 #include<cstdio>
 #include<algorithm>
 using namespace std;

 const long long p=1e9+;
 long long a[],fac[],ans;
 int n,t;

 long long power(long long a,long long b){
     if(a<=)return a;
     if(b==)return ;
     if(b==)return a;
     long long tmp=power(a,b>>);
     tmp=(tmp*tmp)%p;
     if(b&==)tmp=(tmp*a)%p;
     return tmp;
 }

 long long C(int m,int n){
     if(m==)return ;
     if(n<)return ;
     long long tmp=(fac[n]*power(fac[m],p-))%p;
     tmp=(tmp*power(fac[n-m],p-))%p;
     return tmp;
 }

 int main(){
     scanf("%d",&t);
     int i,j,s,k,l;
     fac[]=;
     for(i=;i<=;i++)fac[i]=(fac[i-]*i)%p;
     for(l=;l<=t;l++){
         ans=;
         scanf("%d",&n);
         for(i=;i<=n;i++)scanf("%I64d",&a[i]);
         a[n+]=;
         sort(a+,a++n);
         s=k=;
         long long div=;
         for(i=;i<=n;i++){
             ++k;
             if(a[i]!=a[i+]){
                 long long tmp,ss=;
                 tmp=(fac[s]*fac[k])%p;
                 tmp=(tmp*fac[n-s-k])%p;
                 tmp=(tmp*C(s,n))%p;
                 for(j=;j<=k;j++)ss+=(((a[i]*j)%p)*C(k-j,n-s-j-))%p;
                 ss%=p;
                 ans+=(ss*tmp)%p;
                 ans%=p;
                 div=(div*fac[k])%p;
                 s+=k;
                 k=;
             }
         }
         ans=(ans*power(div,p-))%p;
         printf("Case #%d: %I64d\n",l,ans);
     }
     return ;
 }

AC代码