Hdu2212 dfs （搜索&剪枝or打表）

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其实可以用搜索做。

3628800=10*fact(9) 就是这个dfs数的上限。

总体思想：搜索+剪枝，先搜出所有各位阶乘和的位数和本身位数相同的数，然后再仔细判断。

我觉得练习的时候就尽量不要用直接打表法做了吧。。

#include"stdio.h"#include"stdlib.h"#include"time.h"#include"math.h"#include"iostream"#include"string.h"#include"algorithm"#include"vector"#include"string"#include"queue"#include"map"#define nmax 1000000#define inf 0x7ffffff#define eps 1e-8#define PI 3.1415926535#define PRIME 9999991#define LL long long///---rand---/////double random(){return (double)(rand()%RAND_MAX )/RAND_MAX ;}//int random(int a,int b){return (double)a+random()*(b-a+1);}///----------///inline int maxy(int a,int b){return a>b?a:b;}inline int miny(int a,int b){return a<b?a:b;}inline long long  maxy(long long  a,long long  b){return a>b?a:b;}inline long long  miny(long long  a,long long  b){return a<b?a:b;}using namespace std;map<long long ,int >mp;//LL max_int =2147483647;LL max_int=10000000000000ll;LL fact[20]={1,1,2,6,24,120,720,5040,40320,362880};LL dig[20]={0ll,1ll,10ll,100ll,1000ll,10000ll,100000ll,1000000ll,10000000ll,100000000ll,1000000000ll,10000000000ll,100000000000ll/*12*/,1000000000000ll/*13*/,10000000000000ll/*14*/};bool is_dfs(LL v){    LL rem=v;    LL ans=0;    while(v>0){        ans+=(LL)fact[v%10];        if(ans>rem)return false;        v/=10;    }    if(ans==rem)return true;    return false;}int stk[20];int cnt;LL ans[nmax];void dfs_tranc(int k,LL cursum,int len){    if(k==len){        if(cursum>=dig[len]&&cursum<dig[len+1]&&mp[cursum]==0){//位数是否符合要求            //is a len number.            mp[cursum]++;            if(is_dfs(cursum))            ans[cnt++]=cursum;        }    return ;}    if(cursum+(LL)len-k>=dig[len+1]||(LL)cursum+(LL)(len-k)*fact[9]<dig[len])return ;//剪枝    int i;    for(i=0;i<=9;i++){        //if(k==0&&i==0)continue;        dfs_tranc(k+1,(LL)cursum+fact[i],len);    }}int main(){    int i;    cnt=0;    mp.clear();    //long st=clock();    for(i=1;i<=7;i++){        dfs_tranc(0,0,i);    }        for(i=0;i<cnt;i++){        printf("%I64d\n",ans[i]);    }    //printf("Time:%ld\n",clock()-st);return 0;}