DFS(dfs)

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=2212

DFS

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6322    Accepted Submission(s): 3898

Problem Description

A DFS(digital factorial sum) number is found by summing the factorial of every digit of a positive integer.

For example ,consider the positive integer 145 = 1!+4!+5!, so it's a DFS number.

Now you should find out all the DFS numbers in the range of int( [1, 2147483647] ).

There
is no input for this problem. Output all the DFS numbers in increasing
order. The first 2 lines of the output are shown below.

 

Input

no input

 

Output

Output all the DFS number in increasing order.

 

Sample Output

1
2
......

 

Author

zjt

题解:开始想到10位数,枚举每一位数字肯定会超时,O(10的10次方) 根据枚举的情况,1!+4!+5!和5!+4!+1!是一样的,所以可以从大到小按照顺序枚举出所有情况,即可,这样下一个位置枚举的时候就不可以出现比最后小的元素了,这里注意0的阶乘是1,而且为了避免第一位出现0 的情况,从最后一位向前枚举,最后的输出结果只有4个数,所以可以之间打表输出这4个数即可

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 //#define N 8776444321ll//要在后面加ll 这样不会超int
 #define ll long long
 const ll N = ;
 ll ans[];
 int c ;
 int f[],g[];
 int a[];
 int ck(ll sum)
 {
     for(int i =  ; i <  ; i++) f[i] = ,g[i] = ;
     ll v = sum;
     ll tm =  ;
     while(v)
     {
         tm+=a[v%];
         f[v%]++;
         v/=;
     }
     v = tm;
     while(tm)
     {
         g[tm%]++;
         tm/=;
     }
     for(int i =  ;i <  ; i++) if( f[i]!=g[i] ) return -;
     return v;
 }
 void dfs(int cur, ll sum)
 {
     ll res = ck(sum);
     if(sum>N) return ;
     if(res!=-) ans[c++] = res;
     for(int i = cur ; i >=  ; i-- )
     {
         if(sum==&&i==) continue;
         dfs(i,sum*+i);
     }
 }
 int main()
 {
     a[] = a[] = ;
     for(int i =  ; i <  ; i++)
         a[i] = a[i-]*i;
     c = ;
     dfs(,);
     sort(ans,ans+c);
     for(int i =  ; i < c ; i++)
         printf("%lld\n",ans[i]);
     return ;
 }

其实得到输出结果后可以直接打表:

 #include <cstdio>
 using namespace std;
 int main()
 {
     puts("");
     puts("");
     puts("");
     puts("");
     return ;
 }