LightOJ 1189 - Sum of Factorials(贪心+阶乘)

Description

Given an integer n, you have to find whether it can be expressed as summation of factorials. For given n, you have to report a solution such that

n = x₁! + x₂! + ... + x_n! (x_i < x_j for all i < j)

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10¹⁸).

Output

For each case, print the case number and the solution in summation of factorial form. If there is no solution then print 'impossible'. There can be multiple solutions, any valid one will do. See the samples for exact formatting.

Sample Input

4

7

7

9

11

Sample Output

Case 1: 1!+3!

Case 2: 0!+3!

Case 3: 1!+2!+3!

Case 4: impossible

                                                                                           

这一题题意就是给你一个数，问哪些数的阶乘等于这个数，并按照样例从小到大输出这些数的每一项，其中每一项只可以用一次，如果不存在这些数，就输出impossible。

贪心的题型，每次考虑最大的放入里面，从后往前找。

#include<cstdio>long long p[21]={1,1,2,6,24,120,720,5040,40320,362880,3628800,39916800,479001600,6227020800,87178291200,1307674368000,20922789888000,355687428096000,6402373705728000,121645100408832000,2432902008176640000};int main(){int t,cas=1;scanf("%d",&t);while(t--){int a[20],flag=0; long long  n;scanf("%lld",&n);   for(int i=20;i>=0;i--)    {if(n>=p[i]){n=n-p[i];a[flag++]=i;}    }printf("Case %d: ",cas++);if(n==0){for(int i=flag-1;i>=0;i--){if(i==flag-1)  printf("%d!",a[i]);else printf("+%d!",a[i]);}printf("\n");}else printf("impossible\n"); }  return 0;}