hdu 4249 A Famous Equation （dp 细节）

最新推荐文章于 2020-09-06 09:46:48 发布

utobe67

最新推荐文章于 2020-09-06 09:46:48 发布

阅读量943

点赞数

CC 4.0 BY-SA版权

分类专栏： dp

本文链接：https://blog.youkuaiyun.com/tobewhatyouwanttobe/article/details/40404841

dp 专栏收录该内容

111 篇文章

订阅专栏

本文介绍了一种解决数学等式中部分数字缺失情况的算法，通过动态规划的方法来确定所有可能的解决方案数量。算法适用于含有加法运算符和等于号的等式，其中某些数字被替换为问号。文章详细解释了如何处理前导零问题，并提供了代码实现。通过实例演示了算法如何计算等式的不同解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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: 1


   
    
     Hint
    
There 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?
?+?=?
?+?=??
?????????+?????????=?????????
*/