hdu 3886 - Final Kichiku “Lanlanshu”（数位dp）多校联合

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 … Dn-1} (D0 > 0), if exists such x, y and z, satisfying:

1. Di−1<Di(0<i<=x)
2. Di−1=Di(x<i<=y)
3. Di−1<Di(y<i<=z)
4. Di−1>Di(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 … Dn-1} (D0 > 0), define x0=0, xl=n-1, if exists such x1, x2…xl (x0 < x1 < x2 < … < xl) satisfying:
1. If str[i]=’/’, Dj−1<Dj(xi<j<=xi+1)
2. If str[i]=’-’, Dj−1=Dj(xi<j<=xi+1)
3. If str[i]=’\’, Dj−1>Dj(xi<j<=xi+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： A,B之间有多少个数，满足给定字符串的趋势
思路:dp[cur][pos][last]表示处理到当前为止，满足趋势字符串到pos位置，上一位是last的有多少个

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;const int MOD=1e8;const int maxn=10;string A,B;string st;int stlen,alen;int dp[110][110][10];int dig[110];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',i--;        A[i]-=1;        return A;    }}int check(int x,int y,char op){    if(op=='/')return x<y;    if(op=='-')return x==y;    return x>y;}int dfs(int cur,int e,int z,int pos,int last){    if(cur==alen)return pos==stlen;    if(!e&&!z&&dp[cur][pos][last]!=-1)        return dp[cur][pos][last];    int end=e?dig[cur]:9;    int ans=0;    for(int i=0;i<=end;i++)    {        if(z)ans+=dfs(cur+1,e&&i==end,z&&i==0,0,i);        else        {            if(pos<stlen&&check(last,i,st[pos]))                ans+=dfs(cur+1,e&&i==end,0,pos+1,i);            else if(pos>0&&check(last,i,st[pos-1]))                ans+=dfs(cur+1,e&&i==end,0,pos,i);        }        ans%=MOD;    }    if(!e&&!z)dp[cur][pos][last]=ans;    return ans;}int solve(string a){    memset(dp,-1,sizeof(dp));    int len=a.size();    alen=a.size();    int st=0;    while(a[st]=='0')st++;    for(int i=st;i<len;i++)dig[i]=a[i]-'0';    return dfs(st,1,1,0,0);}int main(){    while(cin>>st)    {        cin>>A>>B;        stlen=st.size();        int len=A.size();        A=subone(A);        printf("%08d\n",(solve(B)-solve(A)+MOD)%MOD);    }    return 0;}