HDU 4722 Good Numbers(找规律+数位DP)

Good Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 610 Accepted Submission(s): 225

Problem DescriptionInputOutputSample InputSample OutputSourceimport java.io.*;import java.util.*;public class Main {BufferedReader bu;PrintWriter pw;long a;long b,num1,num2;int n;public static void main(String[] args) throws IOException{new Main().work();}void work() throws IOException{bu=new BufferedReader(new InputStreamReader(System.in));pw=new PrintWriter(new OutputStreamWriter(System.out),true);n=Integer.parseInt(bu.readLine());for(int p=1;p<=n;p++){String str[]=bu.readLine().split(" ");a=Long.parseLong(str[0]);b=Long.parseLong(str[1]);for(long i=a;;i++){if(fun(i)){num1=i;//寻找可以被10整除的数break;}}for(long i=b;;i--){if(fun(i)){num2=i;//寻找可以被10整除的数break;}}pw.print("Case #"+p+": ");pw.println(num2/10-num1/10+1);}}//判断每一位的和能否被10整除boolean fun(long a){long sum=0;while(a!=0){sum+=a%10;a/=10;}if(sum%10==0)return true;return false;}}

第二种方法

import java.io.*;import java.math.BigInteger;import java.util.*;public class Main {int t;long a,b;PrintWriter pw;long digdit[]=new long[30];long[][] dp=new long[30][30];public static void main(String[] args) throws IOException{new Main().work();}void work() throws IOException{BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));pw=new PrintWriter(new OutputStreamWriter(System.out),true);t=Integer.parseInt(bu.readLine());for(int i=1;i<=t;i++){pw.print("Case #"+i+": ");String str[]=bu.readLine().split(" ");a=Long.parseLong(str[0]);b=Long.parseLong(str[1]);long num1=dit(b);long num2=dit(a-1);long ans=num1-num2;pw.println(ans);}}long dit(long x){int pos=0;for(int i=0;i<dp.length;i++)Arrays.fill(dp[i], -1);while(x!=0){digdit[pos++]=x%10;//将数字的每一位存储到数组中去x=x/10;}return DFS(pos-1,0,true);}long DFS(int pos,int index,boolean boo){if(pos==-1){if(index==0)return 1;elsereturn 0;}if(!boo&&dp[pos][index]!=-1){ return dp[pos][index];}long s=0;int nindex;long e=boo?digdit[pos]:9;for(int i=0;i<=e;i++){nindex=(index+i)%10;//判断是否可以被10整除s=s+(DFS(pos-1,nindex,(boo&&(i==e))));}if(!boo) dp[pos][index]=s;return s;}}