hdu3709（数位dp)

求区间[l,r]内有多少个数的满足：   选一个位为中点，是的左边的数到该位的距离等于右边的位到该位的距离。     

比如4139  选择3位中点， 那么左边的距离是 4 * 2 + 1 * 1 ， 右边的距离是9 * 1

想了半天，想到了枚举哪一位作为中点， 然后进行数位dp， 但是样例错了， 忽然想到会重复啊，就百度了一下

原来只有0在枚举中点时会重复，其他的数有且只有一个中点是的左边的距离等于右边的距离

所以只要最后ans-len+1即可

#include <functional>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <math.h>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#pragma warning(disable:4996)
using namespace std;
typedef long long LL;
const int INF = 0x7fffffff;
const int mod = 2520;
LL dp[22][22][4000];
int num[22];

LL dfs(int cur, int mid, int diff, bool flag)
{
    if (cur == 0) return diff == 2000;
    if (!flag && dp[cur][mid][diff] != -1)
        return dp[cur][mid][diff];
    LL ret = 0;
    int end = flag ? num[cur] : 9;
    for (int i = 0; i <= end; ++i)
    {
        int newDiff = diff;
        if (cur > mid)
            newDiff += (cur - mid) * i;
        else if (cur < mid)
            newDiff -= (mid - cur) * i;
        ret += dfs(cur - 1, mid, newDiff, i == end && flag);
    }
    if (!flag)
        dp[cur][mid][diff] = ret;
    return ret;
}
LL calc(LL n)
{
    int len = 0;
    while (n)
    {
        num[++len] = n % 10;
        n /= 10;
    }
    LL ret = 0;
    
    for (int i = 1; i <= len; ++i)
        ret += dfs(len, i, 2000,true);
    //0每次枚举中点时都被计算进去了，所以要剪掉
    return ret - len + 1;
    
}
int main()
{
    memset(dp, -1, sizeof(dp));
    int t;
    LL l, r;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%I64d%I64d", &l, &r);
        if (l == 0)
            printf("%I64d\n", calc(r));
        else
            printf("%I64d\n", calc(r) - calc(l - 1));
    }
        return 0;
}