HDU - 3709 - Balanced Number（数位DP)

链接：

https://vjudge.net/problem/HDU-3709

题意：

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 42 + 11 = 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].

思路：

刚开始没算具体大小。。以为直接算会暴空间。
记录当前的力矩和，让高位表示正数，这样在从高位到低位的时候，如果力矩边成负值，可以直接返回0，因为低位的权值是负的。
同时要枚举每一个位置作为支点的情况，再去减掉每次算的0.

代码:

// #include<bits/stdc++.h>
#include<iostream>
#include<cstdio>
#include<vector>
#include<string.h>
#include<set>
#include<queue>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
const int MOD = 1e9+7;
const int MAXN = 1e6+10;

LL F[20][20][1600];
LL dig[20];
LL a, b;

LL Dfs(int pos, int piv, int sum, bool lim)
{
    if (pos == -1)
        return sum == 0;
    if (sum < 0)
        return 0;
    if (!lim && F[pos][piv][sum] != -1)
        return F[pos][piv][sum];
    int up = lim ? dig[pos] : 9;
    LL ans = 0;
    for (int i = 0;i <= up;i++)
        ans += Dfs(pos-1, piv, (pos-piv)*i+sum, lim && i == up);
    if (!lim)
        F[pos][piv][sum] = ans;
    return ans;
}

LL Solve(LL x)
{
    int p = 0;
    while(x)
    {
        dig[p++] = x%10;
        x /= 10;
    }
    LL ans = 0;
    for (int i = 0;i < p;i++)
        ans += Dfs(p-1, i, 0, true);
    return ans-(p-1);
}

int main()
{
    // freopen("test.in", "r", stdin);
    memset(F, -1, sizeof(F));
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%lld%lld", &a, &b);
        printf("%lld\n", Solve(b)-Solve(a-1));
    }

    return 0;
}