hdu4722（数位dp）

Good Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3971    Accepted Submission(s): 1269

Problem Description

If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.

 

Input

The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10¹⁸).

 

Output

For test case X, output "Case #X: " first, then output the number of good numbers in a single line.

 

Sample Input

2
1 10
1 20

 

Sample Output

Case #1: 0
Case #2: 1
Hint
The answer maybe very large, we recommend you to use long long instead of int.

题意：求区间[l,r]之间的数的位数和能整除10的数的个数。

思路：数位dp。dp[i][j]表示i位数其中对10取余为j的数的个数。

数位dp有两大阵营，递推派与递归派，各有其长，递归的更容易敲，递推的更容易想。不过递推的要想完全理解真的挺难，所以还是回归递归吧！

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
ll dp[20][10];///dp[i][j]表示i位对10取余为j的有多少个数
int digit[20];
ll dfs(int pos,int st,bool limit)
{
    if(pos==0)return st==0;
    if(!limit&&dp[pos][st]!=-1)return dp[pos][st];
    ll ans=0;
    int end=limit?digit[pos]:9;
    for(int i=0; i<=end; i++)
        ans+=dfs(pos-1,(i+st)%10,limit&&(i==end));
    if(!limit)
        dp[pos][st]=ans;
    return ans;
}
ll get(ll x)
{
    int bj=0;
    while(x)
        digit[++bj]=x%10,x/=10;
    return dfs(bj,0,1);
}
int main()
{
    int t,o=1;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        ll l,r;
        scanf("%I64d%I64d",&l,&r);
        printf("Case #%d: %I64d\n",o++,get(r)-get(l-1));
    }
    return 0;
}

自己以前写的递推：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
ll dp[20][20];

void init()
{
    for(int i=0; i<10; i++)
        dp[1][i]=1;
    for(int i=2; i<=20; i++)
        for(int j=0; j<10; j++)
            for(int k=0; k<10; k++)
                dp[i][(j+k)%10]+=dp[i-1][k];
}

ll get(ll x)
{
    if(x==0)return 1;
    int digit[20],bj=0,sum=0;
    while(x)
        digit[++bj]=x%10,x/=10;
    ll ans=0;
    for(int i=bj; i>0; i--)
    {
        if(i>1)
            for(int j=0; j<digit[i]; j++)
                ans+=dp[i-1][(sum+j)%10];
        else
            for(int j=0; j<=digit[i]; j++)
                if((sum+j)%10==0)
                    ans++;
        sum=(sum+digit[i])%10;
    }
    return ans;
}
int main()
{
    init();
    int t,o=1;
    scanf("%d",&t);
    while(t--)
    {
        ll a,b;
        scanf("%I64d%I64d",&a,&b);
        printf("Case #%d: %I64d\n",o++,get(b)-get(a-1));
    }
    return 0;
}