hdu3709 Balanced Number

Balanced Number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 6039    Accepted Submission(s): 2888

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

 

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题意：区间[l,r]，求其中有多少个数n，满足以一位为支点，另外两边的数字大小乘以力矩之和相等。

这样的数叫做平衡数，题目要求求出区间中平衡数的个数。

这相比最基础的数位DP有一些提高，但也不难。我们枚举以哪一位为支点，然后将两边数字大小乘以力矩和的差求出来，差为0的数即为我们所要求得到的平衡数。

剩下的方法和基础数位DP就是大同小异了。

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>#define ll long longusing namespace std;int digit[20];ll dp[20][20][2005];ll dfs(int len,int pivot,int l,int flag){      if(!len) return l==0;     if(l<0) return 0;      if(!flag && dp[len][pivot][l]!=-1) return dp[len][pivot][l];    ll sum=0;    int top=9;    if(flag) top=digit[len];    for(int i=0;i<=top;i++){        int next=l;        next+=(len-pivot)*i;        sum+=dfs(len-1,pivot,next,flag&&i==top);    }    if(!flag) dp[len][pivot][l]=sum;    return sum;}ll solve(ll x){    int len=0;    while(x){          digit[++len]=x%10;        x/=10;    }    ll ans=0;    for(int i=1;i<=len;i++){        ans+=dfs(len,i,0,1);    }    return ans-len;}int main(){      int T;      ll l,r;    scanf("%d",&T);    memset(dp,-1,sizeof(dp));    while(T--){        scanf("%I64d%I64d",&l,&r);        printf("%I64d\n",solve(r)-solve(l-1));    }    return 0;}