Add Again (repeating element sort)

ADD Again
Input: standard input

Output: Standard Output

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

Input

Each input set would start with a positive integern (1≤n≤12). The next line would contain N decimal digits. Input would be terminated by n=0. There'll is at the most 20000 test set.

Output

For each test set, there should is a one line output containing the summation. The value would 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

/* Test instructions probably is to give you n number, with this n number can compose M integer, for m integer and this problem knowledge point is a repeating element of the total number of elements: there are k elements, including the first element I have NI, perfection permutation number: The number of =n!/(N1!*n2!*n3!*n4!.... *nk!) Figure out each number of times each position, and then add up on the PS: This problem will be a long long to use unsigned long long, seemingly UVA before the problem often has this kind of egg pain ah */

#include <stdio.h> #include <string.h> #define LL unsigned long longint a[10];//The topic is not very clear, is the number between 0-9; 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];    }} int main () {    freopen ("Add.txt", "R", stdin);    LL ANS,SUM,TP;    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+=TP;        }        Ans=jie[n-1]*sum;        for (LL i=0;i<10;i++)        {            if (A[i])            Ans/=jie[a[i]];        }        LL kk=0;        for (int i=1;i<=n;i++)        {            kk=kk*10+ans;        }        printf ("%llu\n", KK);    }    return 0;}

all permutations = (n-1)!/(n1!* N2!*n3!*n4!.. (ni-1)!.. *nk!) Figure out the number of times each position is counted, and then multiply I to add up the contribution value of the I element

#include <stdio.h> #include <string.h> #define LL unsigned long longint a[10];//The topic is not very clear, is the number between 0-9; 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", &AMP;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\n", KK); } return 0;}

Add Again (repeating element sort)