hdu 4249 A Famous Equation （dp 细节）
A Famous Equation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 514 Accepted Submission(s): 153
Problem Description
Mr. B writes an addition equation such as 123+321=444 on the blackboard after class. Mr. G removes some of the digits and makes it look like “1?3+??1=44?”. Here “?” denotes removed digits. After Mr. B realizes some digits are missing, he wants to recover them. Unfortunately, there may be more than one way to complete the equation. For example “1?3+??1=44?” can be completed to “123+321=444” , “143+301=444” and many other possible solutions. Your job is to determine the number of different possible solutions.
Input
Each test case describes a single line with an equation like a+b=c which contains exactly one plus sign “+” and one equal sign “=” with some question mark “?” represent missing digits. You may assume a, b and c are non-negative integers, and the length of each number is no more than 9. In the other words, the equation will contain three integers less than 1,000,000,000.
Output
For each test case, display a single line with its case number and the number of possible solutions to recover the equation.
Sample Input
7+1?=1? ?1+?1=22
Sample Output
Case 1: 3 Case 2: 1HintThere are three solutions for the first case: 7+10=17, 7+11=18, 7+12=19 There is only one solution for the second case: 11+11=22 Note that 01+21=22 is not a valid solution because extra leading zeros are not allowed.
Source
Fudan Local Programming Contest 2012
题意：
给你一个式子a+b=c，有些位变为不确定?，问你有多少种可能使等式成立。
注意不能有前导0，0不算。
思路：
先把a、b都添0位数和c补齐，dp[i][j][k][2]表示到第i位时，a那位数字为j，b那位数字为k，第i位是否有进位的种数。逐位递推，枚举每一位可能的情况，注意前导0的时不要转移。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <string>
#define mod 1000000007
typedef long long ll;
using namespace std;
int n,lena,lenb,lenc;
char s[105];
string a,b,c,aa,bb;
ll ans,dp[15][15][15][2];
void solve()
{
int i,j,k,jj,kk;
memset(dp,0,sizeof(dp));
dp[lenc][0][0][0]=1;
for(i=lenc; i>0; i--)
{
for(j=0; j<=9; j++)
{
if(lenc-i==lena&&lena>1&&j==0) continue ; // 前导0 不要转移
for(k=0; k<=9; k++)
{
if(lenc-i==lenb&&lenb>1&&k==0) continue ; // 前导0 不要转移
if(dp[i][j][k][0]==0&&dp[i][j][k][1]==0) continue ;
for(jj=0; jj<=9; jj++)
{
if(!(a[i-1]=='?'||a[i-1]==jj+'0')) continue ;
for(kk=0; kk<=9; kk++)
{
if(!(b[i-1]=='?'||b[i-1]==kk+'0')) continue ;
if(c[i-1]=='?'||(jj+kk)%10==c[i-1]-'0')
{
if(lenc>1&&i-1==0&&jj+kk==0) ;
else dp[i-1][jj][kk][(jj+kk>=10)?1:0]+=dp[i][j][k][0];
}
if(c[i-1]=='?'||(jj+kk+1)%10==c[i-1]-'0')
{
dp[i-1][jj][kk][(jj+kk+1>=10)?1:0]+=dp[i][j][k][1];
}
}
}
}
}
}
ans=0;
for(j=0;j<=9;j++)
{
if(lena==lenc&&j==0&&lena>1) continue ;
for(k=0;k<=9;k++)
{
if(lenb==lenc&&k==0&&lenb>1) continue ;
ans+=dp[0][j][k][0];
}
}
}
int main()
{
int i,j,ca=0;
while(~scanf("%s",s))
{
a="";
lena=lenb=lenc=0;
for(i=0; s[i]!='+'; i++) lena++,a+=s[i];
b="";
for(i++; s[i]!='='; i++) lenb++,b+=s[i];
c="";
for(i++; s[i]!='\0'; i++) lenc++,c+=s[i];
aa=a; bb=b;
for(i=1; i<=lenc-lena; i++) a="0"+a;
for(i=1; i<=lenc-lenb; i++) b="0"+b;
solve();
printf("Case %d: %I64d\n",++ca,ans);
}
return 0;
}
/*
?+0=?
0+?=?
?+?1=??
1?+?=?1
?+??=???
?+??=10?
?+?=?
?+?=??
?????????+?????????=?????????
*/