HNU1[B题] DP，数位DP

题目链接：http://acm.hnu.cn/online/?action=problem&type=show&id=12813&courseid=267

解题思路：DP，数位DP，比赛时候太蠢写了个187行的DP，结果一看标程37行，给跪。

解题代码：

这是标程

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
using namespace std;
#define LL __int64
const LL M=1e9+7;
char s1[1010],s[1010],s2[1010];
LL dp[100][3];
int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
    while (gets(s))
    {
        memset(dp,0,sizeof(dp));
        if (s[0]=='0') break;
        gets(s1);gets(s2);
        int l=strlen(s);
        dp[l][0]=1;
        for (int i=l-1;i>=0;i--) //   从右到左第几位 
            for (int p=0;p<2;p++)     //  p表示是否进位0,1 
                if (dp[i+1][p]>0)   //表示上一位是否存在
                    for (int j=0;j<10;j++) // 假设s上i位的数字是j 
                    {
                        if (i==0 && j==0) continue;   //最高位不能为0 
                        if (s[i]=='?' || s[i]-'0'==j)
                        for (int k=0;k<10;k++)   //假设s1上的i位数字是k 
                        {
                            if (i==0 && k==0) continue;
                            if (s1[i]=='?' || s1[i]-'0'==k)
                            if (s2[i]=='?' || s2[i]-'0'==(j+k+p)%10)
                                dp[i][(j+k+p)/10]=(dp[i][(j+k+p)/10]+dp[i+1][p])%M;
                        }
                    }
        cout<<dp[0][0]<<endl;
    }
    return 0;
}

下面是我蠢代码

// File Name: b.cpp
// Author: darkdream
// Created Time: 2014年07月19日 星期六 12时59分09秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long

using namespace std;

char a[200];
char b[200];
char c[200];
int ia[200];
int ib[200];
int ic[200];
int len ;
void change()
{
   for(int i= 0 ;i < len ;i ++)
   {
       if(a[i] == '?')
       {
          ia[len-i] = 11;    
       }else{
          ia[len-i] = a[i]-'0';
       }
   }
   for(int i= 0 ;i < len ;i ++)
   {
       if(b[i] == '?')
       {
          ib[len-i] = 11;    
       }else{
          ib[len-i] = b[i]-'0';
       }
   }
   for(int i= 0 ;i < len ;i ++)
   {
       if(c[i] == '?')
       {
          ic[len-i] = 11;    
       }else{
          ic[len-i] = c[i] -'0';
       }
   }

}
LL sdp[12][12][12];
LL sdp1[12][12][12];
LL adp[12][12][12];
LL adp1[12][12][12];
void init()
{
  memset(sdp,0,sizeof(sdp));
  memset(adp,0,sizeof(adp));
  memset(adp1,0,sizeof(adp1));
  memset(sdp1,0,sizeof(sdp1));
  for(int i = 0 ;i <= 9 ;i ++)
     for(int j = 0 ;j <= 9 ;j ++)
     {
        if(i +j <= 9)
        {
           int t = i + j ;
           sdp[11][11][11] ++;
           sdp[11][11][j] ++;
           sdp[11][i][11] ++;
           sdp[11][i][j] ++;
           sdp[t][11][11] ++ ;
           sdp[t][11][j] ++;
           sdp[t][i][11] ++;
           sdp[t][i][j]++;
        }
        if(i + j > 9) 
        {
           int t = (i+j) % 10 ;
           sdp1[11][11][11] ++;
           sdp1[11][11][j] ++;
           sdp1[11][i][11] ++;
           sdp1[11][i][j] ++;
           sdp1[t][11][11] ++ ;
           sdp1[t][11][j] ++;
           sdp1[t][i][11] ++;
           sdp1[t][i][j]++;
        }
        if(i+j+1 <= 9)
        {
           int t = i + j + 1;
           adp[11][11][11] ++;
           adp[11][11][j] ++;
           adp[11][i][11] ++;
           adp[11][i][j] ++;
           adp[t][11][11] ++ ;
           adp[t][11][j] ++;
           adp[t][i][11] ++;
           adp[t][i][j]++;
        }
        if(i+j+1 > 9){
           int t = (i+j+1) % 10 ;
           adp1[11][11][11] ++;
           adp1[11][11][j] ++;
           adp1[11][i][11] ++;
           adp1[11][i][j] ++;
           adp1[t][11][11] ++ ;
           adp1[t][11][j] ++;
           adp1[t][i][11] ++;
           adp1[t][i][j]++;
        }
     }
}
LL dp[200][3];
#define M 1000000007
LL lenans(int a, int b, int c,int t)
{
   LL sum = 0 ;
  // printf("%d %d %d %d
",a,b,c,t);
   for(int i = 1;i <= 9;i ++)
   {
       if(i == a || a == 11)
       for(int j = 1;j <= 9 ;j ++ )
       {
           if((j == b || b == 11))
           {   if(i +j + t<= 9 && (i+j+t == c || c == 11))
                    sum ++ ;  
          
    //            printf("%d %d %d %d
",i,j,i+j+t,c);
           }
       }
   }
  // printf("%lld
",sum);
   return sum;  
}
int main(){ 
  init();
  while(scanf("%s",a) != EOF)
  {
    if( a[0] == '0' )
        break;
    scanf("%s %s",b,c);
    len = strlen(a);    
    change();
    // 1 not
    // 2 yes
//    for(int i =1 ;i <= len ;i ++)
//        printf("%d ",ic[i]);
//    printf("
");
    memset(dp,0,sizeof(dp));
    dp[0][1] = 1; 
    dp[0][2] = 0; 
    for(int i = 1;i <= len ;i ++)
    {
       int ta,tb,tc;
       tc = ic[i];
       ta = ia[i];
       tb = ib[i];
       if(i != len )
       {
         dp[i][1] = (dp[i-1][1]*sdp[tc][ta][tb] % M + dp[i-1][2]*adp[tc][ta][tb] %M) % M;
         dp[i][2] = (dp[i-1][1]*sdp1[tc][ta][tb] % M + dp[i-1][2]*adp1[tc][ta][tb] %M) % M;
       }else{
         dp[i][1] = (dp[i-1][1]*lenans(ta,tb,tc,0) % M + dp[i-1][2]*lenans(ta,tb,tc,1)% M)% M; 
       }
//      printf("%lld %lld
",dp[i][1],dp[i][2]); 
    }
    printf("%lld
",dp[len][1]);
//    printf("%lld",sdp[11][11][11]);
  }

return 0;
}