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+7;
long long a[100100],fac[100100],ans;
int n,t;

long long power(long long a,long long b){
    if(a<=1)return a;
    if(b==0)return 1;
    if(b==1)return a;
    long long tmp=power(a,b>>1);
    tmp=(tmp*tmp)%p;
    if(b&1==1)tmp=(tmp*a)%p;
    return tmp;
}

long long C(int m,int n){
    if(m==0)return 1;
    if(n<0)return 0;
    long long tmp=(fac[n]*power(fac[m],p-2))%p;
    tmp=(tmp*power(fac[n-m],p-2))%p;
    return tmp;
}

int main(){
    scanf("%d",&t);
    int i,j,s,k,l;
    fac[0]=1;
    for(i=1;i<=100000;i++)fac[i]=(fac[i-1]*i)%p;
    for(l=1;l<=t;l++){
        ans=0;
        scanf("%d",&n);
        for(i=1;i<=n;i++)scanf("%I64d",&a[i]);
        a[n+1]=0;
        sort(a+1,a+1+n);
        s=k=0;
        long long div=1;
        for(i=1;i<=n;i++){
            ++k;
            if(a[i]!=a[i+1]){
                long long tmp,ss=0;
                tmp=(fac[s]*fac[k])%p;
                tmp=(tmp*fac[n-s-k])%p;
                tmp=(tmp*C(s,n))%p;
                for(j=1;j<=k;j++)ss+=(((a[i]*j)%p)*C(k-j,n-s-j-1))%p;
                ss%=p;
                ans+=(ss*tmp)%p;
                ans%=p;
                div=(div*fac[k])%p;
                s+=k;
                k=0;
            }
        }
        ans=(ans*power(div,p-2))%p;
        printf("Case #%d: %I64d\n",l,ans);
    }
    return 0;
}