Uva 11076 Add Again 解题报告（组合数学）

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 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

Problemsetter: Md. Kamruzzaman

Special Thanks: Shahriar Manzoor

    解题报告：假设不去重，那么n个数一共有n!种组合。以第一位为例，n个数各出现(n-1)!次，那么第一位的和就是(n-1)!*sum。再去重，每个数出现p次，结果就除以p!。最后统计成一个数字即可。代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef long long LL;
LL factorial[20];

void init()
{
factorial[0]=1;
for(int i=1;i<=12;i++)
factorial[i]=factorial[i-1]*i;
}

int d[10];
int main()
{
init();

int n;
while(~scanf("%d",&n) && n)
{
memset(d,0,sizeof(d));

LL ans = factorial[n-1];
int sum = 0;
for(int i=0;i<n;i++)
{
int tmp;
scanf("%d",&tmp);
d[tmp]++;
sum+=tmp;
}
ans*=sum;

for(int i=0;i<10;i++)
ans/=factorial[d[i]];

LL res=0;
for(int i=0;i<n;i++)
res = res*10+ans;

printf("%lld\n", res);
}
}