Add Again(重复元素排序)

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 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

 

/*题意大概就是给你n个数,用这n个数能组成m个整数，求m个整数的和这题知识点就一个就是有重复元素的全排列个数: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!/(n1!*n2!*n3!*n4!....*nk!)算出每个数出现在每个位置的次数，然后加起来就可以了ps:这道题会卡long long 要用unsigned long long ,貌似uva以前的题经常有这种蛋疼的问题啊*/

 

#include<stdio.h>#include<string.h>#define LL unsigned long longint 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];    }}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;}

 

 

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

#include<stdio.h>#include<string.h>#define LL unsigned long longint 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\n",kk);    }    return 0;}