UVa 11076 - Add Again （排列之和 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 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

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Problemsetter: Md. Kamruzzaman

Special Thanks: Shahriar Manzoor

 

 

题意：

输入ｎ个数字，这些数字的任何一种排列都是一个整数，你的任务是求出所有这些整数之和。

对于ｍ个数，要求他的和，可以像这样求

对于每一位ｉ，求ｍ个数第ｉ为上的和然后乘第ｉ位相应的权值10^i（ｉ从０开始）

对于第ｉ位ｊ出现的方法数则是除去一个ｊ后的其他数的排列方法数

这个题unsigned long long 也会超

但是 unsigned long long / long long 也能过

数据没有这么大？

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64d\n"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lld\n"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform, &(a))
#define P64I(a) printf(oform, (a))
#define P64I1(a) printf(oform1, (a))
#define REP(i, n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 10e-9;
const double PI = (4.0*atan(1.0));

int vis[10];
unsigned long long C[20][20];

void init(int n) {
    memset(C, 0, sizeof(C));
    C[0][0] = 1;
    for(int i=1; i<=n; i++) {
        C[i][0] = C[i][i] = 1;
        for(int j=1; j<i; j++) {
            C[i][j] = C[i-1][j-1] + C[i-1][j];
        }
    }
}

int main() {
    int n;

    init(15);
    while(scanf("%d", &n) != EOF && n) {
        int sum = 0;
        memset(vis, 0, sizeof(vis));
        for(int i=0; i<n; i++) {
            int t;
            scanf("%d", &t);
            vis[t]++;
            sum += t;
        }
        unsigned long long p10 = 1;
        unsigned long long ans = 0;
        for(int i=0; i<n; i++) {
            for(int j=1; j<10; j++) if(vis[j]) {
                vis[j]--;
                int lev = n-1;
                LL tt = 1;
                for(int k=1; k<10; k++) if(vis[k]) {
                    tt *= C[lev][vis[k]];
                    lev -= vis[k];
                }
                ans += p10 * j * tt;
                vis[j]++;
            }
            p10 *= 10;
        }
        cout<<ans<<endl;
    }

    return 0;
}