ZJUT Practice 3.10 MAYBE JIANDAN! G - Add Again（uva 11107）

Problem C Add Again Input: Standard Input

Output: Standard Output

Summation of sequence of integers is always a common problem in Computer Science. Rather than computing blindly, some intelligent techniques make the task simpler. Here you have to find the summation of a sequence of integers. The sequence is an interesting one and it is the all possible permutations of a given set of digits. For example, if the digits are <1 2 3>, then six possible permutations are <123>, <132>, <213>, <231>, <312>, <321> and the sum of them is 1332.

Input

Each input set will start with a positive integer N (1≤N≤12). The next line will contain N decimal digits. Input will be terminated by N=0. There will be at most 20000 test set.

Output

For each test set, there should be a one line output containing the summation. The value will fit in 64-bit unsigned integer.

Sample Input                             Output for Sample Input

3 1 2 3 3 1 1 2 0

1332 444

 

Problemsetter: Md. Kamruzzaman

Special Thanks: Shahriar Manzoor

 

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
int a[10],permu[13]={1};
int main(){
    for(int i=1;i<=12;i++)
        permu[i]=i*permu[i-1];
    for(int n;scanf("%d",&n) && n;){
        memset(a,0,sizeof(a));
        int temp;
        for(int i=0;i<n && scanf("%d",&temp);i++)
            a[temp]++;
        long long int ans=0;
        for(int i=1;i<=9;i++){
            long long int sum=1;
            if(a[i]>0){
                for(int j=0;j<=9;j++){
                    if(j==i)
                        sum*=permu[a[j]-1];
                    else
                        sum*=permu[a[j]];
                }
                ans+=i*permu[n-1]*1.0/sum;
            }
        }
        long long int final_ans=0;
        for(int i=0;i<n;i++)
            final_ans=final_ans*10+ans;
        printf("%lld\n",final_ans);
    }
}

final_ans=111.....1(n位1)*ans;

ans=∑i*(n-1)!/(a1!*a2!*....(ai-1)!*...*a9!) ;(i=0-->9)

公式就是这样。