UVa 11076 Add Again (组合数学)

11076 - Add Again

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=467&page=show_problem&problem=2017

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

思路：平均数思想

由于每个数出现在各个位的次数是一样的，

所以ans=每个位的平均数*排列数*n个1

比如<1 1 2 2>，我们有：

完整代码：

/*0.042s*/

#include<cstdio>
#include<cstring>
const long long one[13] =
{
	0, 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111,
	111111111, 1111111111, 11111111111, 111111111111
};

long long a[10], fac[13];//factorial

int main(void)
{
	int n, num, count;
	long long ans;
	fac[0] = 1;
	for (int i = 1; i <= 12; i++)
		fac[i] = i * fac[i - 1];///计算阶乘
	while (scanf("%d", &n), n)
	{
		memset(a, 0, sizeof(a));
		count = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &num);
			count += num;
			a[num]++;
		}
		ans = fac[n - 1] * count;
		for (int i = 0; i < 10; ++i)
			ans /= fac[a[i]];
		printf("%lld\n", ans * one[n]);
	}
}