HDU

Final Kichiku “Lanlanshu”

Problem Description

During 2010 summer training, temperlisyer often does problem like this:
“Consider a decimal integer as sequence of digits {D₀, D₁ … D_n-1} (D₀ > 0), if exists such x, y and z, satisfying:
1.D_i-1<D_i (0<i<=x)
2.D_i-1=D_i (x<i<=y)
3.D_i-1<D_i (y<i<=z)
4.D_i-1>D_i (z<i<=n-1)

We call this integer “Lanlanshu”, now give you two numbers A and B, calculate how many “Lanlanshu” are in [A, B].“
He solved so many of these and finally get bored, and then get crazy! He decided to make up a problem to put this type of problems to an end.
Give you a string str consists only by ‘/’, ‘-‘ and ‘\’, and its length is l. Consider a decimal integer as sequence of digits {D₀, D₁ … D_n-1} (D0 > 0), define x₀=0, x_l=n-1, if exists such x₁, x₂...x_l (x₀ < x₁ < x₂ < ... < x_l) satisfying:
1. If str[i]=’/’, D_j-1<D_j (x_i<j<=x_i+1)
2. If str[i]=’-’, D_j-1=D_j (x_i<j<=x_i+1)
3. If str[i]=’\’, D_j-1>D_j (x_i<j<=x_i+1)

We call it Final Kichiku “Lanlanshu”, now give you two numbers A and B, calculate how many Final Kichiku “Lanlanshu” are in [A, B]. This number maybe huge, we only want to now the last 8 digits of the result.

 

Input

Multiple cases (no more than 100), for each case:
The first line is string str, length is below 100.
The second line contains two integers A and B (0≤Ai≤Bi≤10^100).
Input terminates by EOF.

 

Output

For each case, output 8 digits representing the last 8 digits of the number of Final Kichiku “Lanlanshu” in [A, B]. If it’s less than 8 digits, fill it with leading zeros.

 

Sample Input

/\01221 2012

 

Sample Output

00000315

 

Author

temperlsyer

 

Source

2011 Multi-University Training Contest 5 - Host by BNU

 

题意：给你一个模式串，找到区间内符合该模式的数字的个数。

解题思路：数位DP，好题！以后类似的题直接套这个模板就行了！代码有详细注释，很好理解。

#include<iostream>#include<deque>#include<memory.h>#include<stdio.h>#include<map>#include<string.h>#include<algorithm>#include<vector>#include<math.h>#include<stack>#include<queue>#include<set>#include<bitset>using namespace std;typedef long long int ll;const int MOD=100000000;int dp[110][110][10];//第i位，前j个字符已经符合，前一个数是k的个数int dig[200];string str;//模式串//去除前导0string qu(string a){    string ans;    bool have=0;    for(int i=0;i<a.size();i++){        if(!have&&a[i]=='0')            continue;        have=1;        ans.push_back(a[i]);    }    return ans;}//字符串减一操作string subone(string a){    int len=a.size();    if(a[len-1]>'0'){        a[len-1]-=1;        return a;    }    else{        int i=len-1;        while(i>=0&&a[i]<='0')            a[i--]='9';        a[i]-=1;        return a;    }}//判断两个字符是否符合当前位的模式串int ok(int x,int y,char op){    if(op=='/')return x<y;    if(op=='-')return x==y;    return x>y;}//当前处理到第pos个字符，模式串匹配到了len，前一个数字是last，是否前导0int dfs(int pos,int len,int last,bool have,bool limit){        if(pos==0)//要判断模式串是否也匹配完        return len==str.size();        //记忆化搜索    if(!limit && !have && dp[pos][len][last]!=-1){        return dp[pos][len][last];    }        int end=limit?dig[pos]:9;    int ans=0;        for(int i=0;i<=end;i++){        if(have){            ans+=dfs(pos-1,0,i,have&&i==0,limit&&i==end);//如果有前导0        }        else{                        if(len<str.size()&&ok(last,i,str[len]))                ans+=dfs(pos-1,len+1,i,have&&i==0,limit&&i==end);//贪心的往后匹配，防止////\这种串的干扰            else{                if(len>0&&ok(last,i,str[len-1]))                    ans+=dfs(pos-1,len,i,have&&i==0,limit&&i==end);//往后不符合，再看看现在的符不符合            }                    }            }    ans%=MOD;        if(!limit&&!have)        dp[pos][len][last]=ans;        return ans;    }int solve(string x){        int pos=0;        for(pos=0;pos<x.size();pos++)        dig[x.size()-pos]=x[pos]-'0';        return dfs(pos,0,0,1,1);}int main(){        while(cin>>str){        memset(dp,-1,sizeof(dp));        string l,r;        cin>>l>>r;        l=qu(l);        r=qu(subone(r));                printf("%08d\n",(solve(r)-solve(l)+MOD)%MOD);    }        return 0;}