11076 - Add Again
Time limit: 3.000 seconds
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]);
}
}
