Add Again(重复元素排序) UVA11076

Add Again

 

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 integerN (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

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Problemsetter: Md. Kamruzzaman

Special Thanks: Shahriar Manzoor

思路：对于第x位，一共有k个元素,其中第i个元素有ni个,求全排列个数：全排列个数=（n-1）!/(n1!*n2!*n3!*n4!..(ni-1)!..*nk!)算出每个数出现在每个位置的次数，然后乘以i加起来就是i元素的贡献值了

转载请注明出处：寻找&星空の孩子

#include<stdio.h>
#include<string.h>
#define LL unsigned long long

int a[10];// 题目没有说很清楚，是0-9之间的数；
int b[15];
LL jie[15];
int N;
void init()
{
    jie[0]=1;
    for(LL i=1;i<15;i++)
    {
        jie[i]=i*jie[i-1];
    }
}
LL chsort(LL x)
{
    LL cnt=1;
    for(int i=0;i<10;i++)
    {
        if(a[i])
        {
            if(i==x)
                cnt*=jie[a[i]-1];
            else
                cnt*=jie[a[i]];
        }
    }
    return cnt;
}
int main()
{
   // freopen("Add.txt","r",stdin);

    LL ans,sum;
    init();
    while(scanf("%d",&N),N)
    {
        sum=0;
        memset(a,0,sizeof(a));
        for(int i=0;i<N;i++)
        {
            int tp;
            scanf("%d",&tp);
            a[tp]++;
 //           sum+=b[i];
        }
        ans=0;
 //       ans=jie[N-1]*sum;
        for(LL i=0;i<10;i++)
        {
            if(a[i])
            {
                ans+=jie[N-1]*i/chsort(i);
            }
        }
        LL kk=0;
        for(int i=1;i<=N;i++)
        {
            kk=kk*10+ans;
        }
        printf("%llu
",kk);
    }
    return 0;
}