Balanced Number（HDU-3709）

最新推荐文章于 2022-04-05 09:12:07 发布

Alex_McAvoy

最新推荐文章于 2022-04-05 09:12:07 发布

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分类专栏： # HDU 动态规划——数位 DP

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动态规划——数位 DP

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本文介绍了一种通过动态规划算法解决平衡数计数问题的方法。平衡数是指可以在某个位上找到支点，使得该位左侧所有数乘以其到支点的距离之和等于右侧对应和的非负整数。文章提供了一个C++实现示例，用于计算给定区间内平衡数的数量。

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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

2
0 9
37604 24324

Sample Output

10
897

题意：给出区间 [l,r] 内平衡数的个数，平衡数是指，以某位作为支点，此位左边的数字*距离的和与右边相等

思路：以 dp[pos][x][sta] 代表第pos位的数支点是x，目前力矩为sta的数的个数，对于每一位，以此位为支点计算力矩，然后判断这个数是否为平衡数

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 21
#define MOD 10007
#define E 1e-6
typedef long long LL;
using namespace std;
int bit[N];
LL dp[N][N][N*100];
LL dfs(int pos,int x,int sta,bool limit)
{
    if(pos<=0)
        return sta==0;
    if(sta<0)
        return 0;
    if(!limit&&dp[pos][x][sta]!=-1)
        return dp[pos][x][sta];

    int res=0;
    int up=limit?bit[pos]:9;
    for(int i=0;i<=up;i++)
    {
        int next=sta+i*(pos-x);
        res+=dfs(pos-1,x,next,limit&&(i==up));
    }
    if(!limit)
        dp[pos][x][sta]=res;

    return res;
}
LL solve(LL x)
{
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }

    LL sum=0;
    for(int i=1;i<=len;i++){
        sum+=dfs(len,i,0,true);
    }
    return sum-(len-1);//减去 00、000 、0000...的情况
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        LL left,right;
        scanf("%lld%lld",&left,&right);

        memset(dp,-1,sizeof(dp));
        printf("%lld\n",solve(right)-solve(left-1));
    }
    return 0;
}