lightoj-1189

1189 - Sum of Factorials
PDF (English) Statistics Forum
Time Limit: 0.5 second(s) Memory Limit: 32 MB
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! + ... + xn! (xi < xj 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
Output for Sample Input
4
7
7
9
11
Case 1: 1!+3!
Case 2: 0!+3!
Case 3: 1!+2!+3!
Case 4: impossible

解题思路:刚开始用dfs 跑，但是爆了，研究下发现，每次的答案出了0！和1！这两个相等的可以替换，其他都是唯一的， 因为阶乘相邻的差距太多，例如n前面的加起来都没有n!大，所以都是必须取的。 然后用这种贪心的策略跑一下就可以了

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

typedef long long ll;
ll arr[20];
int save[20];

void init(){
    arr[0] = arr[1] = 1;
    for(int i=2;i<=20;i++){
        arr[i] = i*arr[i-1];//cout<<arr[i]<<endl;
    }
    return ;
}


int main(){
    init();
    int T;
    ll n;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        scanf("%lld",&n);
        printf("Case %d: ",t);
        for(int i=0;i<=20;i++){
            if(n-arr[i]<0){
                printf("impossible
");
                break;
            }
            int ans = 0;
            ll nn = n;
            for(int j=i;j>=0;j--){                
                if(nn-arr[j]>=0){
                    nn -= arr[j];
                    save[ans++] = j;
                    //cout<<nn<<endl;
                    if(nn==0){
                        printf("%d!",save[ans-1]);
                        for(int k=ans-2;k>=0;k--) printf("+%d!",save[k]);
                        printf("
");
                    }
                }                
            }
            if(nn==0) break;
        }
        
    } 
    return 0;
}

//bool dfs(ll num,int t,int t1){
//    if(num==0){
//        printf("%d!",save[0]);
//        for(int i=1;i<t1;i++){
//            printf("+%d!",save[i]);
//        }
//        printf("
");
//        return true;
//    }
//    if(num<0) return false;
//    for(int i=t;arr[i]<=num&&i<20;i++){
//        save[t1] = i;//cout<<i<<endl;//cout<<(num-arr[i])<<" "<<arr[i]<<endl;
//        if(dfs(num-arr[i],i+1,t1+1)==true) return true;
//        
//    }
//    return false;
//}