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 = 8776444321;
ll ans[500];
int c ;
int f[10],g[10];
int a[10];
int ck(ll sum)
{
    for(int i = 0 ; i < 10 ; i++) f[i] = 0,g[i] = 0;
    ll v = sum;
    ll tm = 0 ;
    while(v)
    {
        tm+=a[v%10];
        f[v%10]++;
        v/=10;
    }
    v = tm;
    while(tm)
    {
        g[tm%10]++;
        tm/=10;
    }
    for(int i = 0 ;i < 10 ; i++) if( f[i]!=g[i] ) return -1;
    return v;
}
void dfs(int cur, ll sum)
{
    ll res = ck(sum);
    if(sum>N) return ;
    if(res!=-1) ans[c++] = res; 
    for(int i = cur ; i >= 0 ; i-- )
    {
        if(sum==0&&i==0) continue;
        dfs(i,sum*10+i);
    }
}
int main()
{
    a[0] = a[1] = 1;
    for(int i = 2 ; i < 10 ; i++)
        a[i] = a[i-1]*i;
    c = 0;
    dfs(9,0);
    sort(ans,ans+c);
    for(int i = 1 ; i < c ; i++)
        printf("%lld
",ans[i]);
    return 0;
}

其实得到输出结果后可以直接打表:

#include <cstdio>
using namespace std;
int main()
{
    puts("1");
    puts("2");
    puts("145");
    puts("40585");
    return 0;
}