hdu3709---Balanced Number(디지털 dp)

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 7604 24324
Sample Output
10 897
Author GAO, Yuan
Source 2010 Asia Chengdu Regional Contest
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dp[i][j][k]는 i자리수를 나타내고 축은 j에 있고 k의 개수를 나타낸다. 그리고 j, 디지털 dp를 열거한다. 0은 매번 계산되고 마지막에 빼야 한다(len-1).

/************************************************************************* > File Name: hdu3709.cpp > Author: ALex > Mail: [email protected] > Created Time: 2015 03 01      18 42 38  ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

LL dp[40][40][4000];
int bit[40];

LL dfs (int cur, int pro, int sum, bool flag, bool zero)
{
    if (cur == -1)
    {
        if (zero)
        {
            return 1;
        }
        return sum == 0;
    }
    if (!flag && ~dp[cur][pro][2000 - sum])
    {
        return dp[cur][pro][2000 - sum];
    }
    LL ans = 0;
    int end = flag ? bit[cur] : 9;
    for (int i = 0; i <= end; ++i)
    {
        if (zero && !i)
        {
            ans += dfs (cur - 1, pro, 0, flag && (i == end), 1);
        }
        else if (zero && i)
        {
            ans += dfs (cur - 1, pro, i * (cur - pro), flag && (i == end), 0);
        }
        else
        {
            ans += dfs (cur - 1, pro, sum + i * (cur - pro), flag && (i == end), 0);
        }
    }
    if (!flag)
    {
        dp[cur][pro][2000 - sum] = ans;
    }
    return ans;
}

LL calc (LL n)
{
    int cnt = 0;
    while (n)
    {
        bit[cnt++] = n % 10;
        n /= 10;
    }
    LL ans = 0;
    for (int i = 0; i < cnt; ++i)
    {
        ans += dfs (cnt - 1, i, 0, 1, 1);
    }
    ans -= (cnt - 1);
    return ans;
}

int main ()
{
    memset (dp, -1, sizeof(dp));
    int t;
    scanf("%d", &t);
    while (t--)
    {
        LL l, r;
        scanf("%I64d%I64d", &l, &r);
        printf("%I64d
", calc (r) - calc (l - 1));
    }
    return 0;
}