hdu3709 - Balanced Number（2010 Asia Chengdu Regional Contest）数位dp

Balanced Number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2720    Accepted Submission(s): 1242

Problem Description

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job
to calculate the number of balanced numbers in a given range [x, y].

 

Input

The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10¹⁸).

 

Output

For each case, print the number of balanced numbers in the range [x, y] in a line.

 

Sample Input

20 97604 24324

 

Sample Output

10897

 

Author

GAO, Yuan

 

Source

2010 Asia Chengdu Regional Contest

题意:求出[x,y]之间的平衡数，所谓的平衡数，就是以其中一个数为中点，两边以杠杆原理一样的平衡

思路：数位dp，dp[cur[mid][l]表示到当前位置，以mid为中点，和为l的有多少

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int maxn=20;int dig[maxn];LL dp[maxn][maxn][2010];LL x,y;LL dfs(int cur,int mid,int l,int e){    if(cur<0)return l==0;    if(l<0)return 0;    if(!e&&dp[cur][mid][l]!=-1)        return dp[cur][mid][l];    LL ans=0;    int end=e?dig[cur]:9;    for(int i=0;i<=end;i++)    {        int tmp=l;        tmp+=(cur-mid)*i;        ans+=dfs(cur-1,mid,tmp,e&&i==end);    }    if(!e)dp[cur][mid][l]=ans;    return ans;}LL solve(LL n){    memset(dp,-1,sizeof(dp));    int len=0;    while(n)    {        dig[len++]=n%10;        n/=10;    }    LL ans=0;    for(int i=0;i<len;i++)        ans+=dfs(len-1,i,0,1);    return ans-len+1;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%I64d%I64d",&x,&y);        printf("%I64d\n",solve(y)-solve(x-1));    }    return 0;}