HDU-4249-A Famous Equation(DP)

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最新推荐文章于 2025-05-29 20:25:16 发布

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分类专栏：动态规划

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本文探讨了一种解决数学等式中缺失数字的编程算法，具体描述了一个等式，其中部分数字被替换为问号。通过动态规划的方法，文章详细解释了如何计算并确定所有可能的解决方案的数量。此算法适用于处理任意长度不超过9的等式，并且能够准确处理前导零、进位以及其他边界情况。

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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

思路：dp[i][0]表示从右往左第i位不进位的方法数，dp[i][1]表示进位的方法数。坑点：①long long。②前导零。③1+1=2这种情况结果为1，加特判还WA了。④细节。

代码写得太挫，还是贴上来吧。。。

#include<stdio.h>
#include<string.h>

char a[50],b[50],c[50],s[100];
long long dp[10][2];

int main()
{
    int la,lb,lc,i,j,k,now,p,q,jp,jq,cas=1;

    while(~scanf("%s",s))
    {
        for(i=0;s[i];i++) if(s[i]=='+' || s[i]=='=') s[i]=' ';

        sscanf(s,"%s%s%s",a,b,c);

        la=strlen(a)-1;
        lb=strlen(b)-1;
        lc=strlen(c)-1;

        dp[0][0]=1;
        dp[0][1]=0;

        now=1;

        while(la>=0 || lb>=0 || lc>=0)
        {
            p=q=jp=jq=0;

            if(la>=0 && a[la]=='?')
            {
                for(i=0;i<=9;i++)
                {
                    if(!la && now>1 && !i) continue;

                    if(lb>=0 && b[lb]=='?')
                    {
                        for(j=0;j<=9;j++)
                        {
                            if(!lb && now>1 && !j) continue;

                            if(lc>=0 && c[lc]=='?')
                            {
                                for(k=0;k<=9;k++)
                                {
                                    if(!lc && now>1 && !k) continue;

                                    if(i+j==k) p++;
                                    if(i+j==k-1) jp++;
                                    if(i+j-10==k) q++;
                                    if(i+j-10==k-1) jq++;
                                }
                            }
                            else
                            {
                                if(lc>=0) k=c[lc]-'0';
                                else k=0;

                                if(i+j==k) p++;
                                if(i+j==k-1) jp++;
                                if(i+j-10==k) q++;
                                if(i+j-10==k-1) jq++;
                            }
                        }
                    }
                    else
                    {
                        if(lb>=0) j=b[lb]-'0';
                        else j=0;

                        if(lc>=0 && c[lc]=='?')
                        {
                            for(k=0;k<=9;k++)
                            {
                                if(!lc && now>1 && !k) continue;

                                if(i+j==k) p++;
                                if(i+j==k-1) jp++;
                                if(i+j-10==k) q++;
                                if(i+j-10==k-1) jq++;
                            }
                        }
                        else
                        {
                            if(lc>=0) k=c[lc]-'0';
                            else k=0;

                            if(i+j==k) p++;
                            if(i+j==k-1) jp++;
                            if(i+j-10==k) q++;
                            if(i+j-10==k-1) jq++;
                        }
                    }
                }
            }
            else
            {
                if(la>=0) i=a[la]-'0';
                else i=0;

                if(lb>=0 && b[lb]=='?')
                {
                    for(j=0;j<=9;j++)
                    {
                        if(!lb && now>1 && !j) continue;

                        if(lc>=0 && c[lc]=='?')
                        {
                            for(k=0;k<=9;k++)
                            {
                                if(!lc && now>1 && !k) continue;

                                if(i+j==k) p++;
                                if(i+j==k-1) jp++;
                                if(i+j-10==k) q++;
                                if(i+j-10==k-1) jq++;
                            }
                        }
                        else
                        {
                            if(lc>=0) k=c[lc]-'0';
                            else k=0;

                            if(i+j==k) p++;
                            if(i+j==k-1) jp++;
                            if(i+j-10==k) q++;
                            if(i+j-10==k-1) jq++;
                        }
                    }
                }
                else
                {
                    if(lb>=0) j=b[lb]-'0';
                    else j=0;

                    if(lc>=0 && c[lc]=='?')
                    {
                        for(k=0;k<=9;k++)
                        {
                            if(!lc && now>1 && !k) continue;

                            if(i+j==k) p++;
                            if(i+j==k-1) jp++;
                            if(i+j-10==k) q++;
                            if(i+j-10==k-1) jq++;
                        }
                    }
                    else
                    {
                        if(lc>=0) k=c[lc]-'0';
                        else k=0;

                        if(i+j==k) p++;
                        if(i+j==k-1) jp++;
                        if(i+j-10==k) q++;
                        if(i+j-10==k-1) jq++;
                    }
                }
            }

            dp[now][0]=dp[now-1][0]*p+dp[now-1][1]*jp;
            dp[now][1]=dp[now-1][0]*q+dp[now-1][1]*jq;

            now++;

            la--;
            lb--;
            lc--;
        }

        printf("Case %d: %I64d\n",cas++,dp[now-1][0]);
    }
}