HDU3709

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3709

ull dfs(int len, int piovt, int sum, bool limit)，目前處理到第len位，第piovt 爲支點，目前的和爲sum，是否受到限制。

枚舉每一位作爲支點，累加  dfs(len, i, 0, true)

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

typedef long long ull;

ull dp[20][20][2005];
int digit[20];

ull dfs(int len, int piovt, int sum, bool limit){
    if (len == 0){
        return sum == 0;
    }
    if (sum < 0){
        return 0;
    }
    if (!limit && dp[len][piovt][sum] != -1){
        return dp[len][piovt][sum];
    }

    int Max = limit ? digit[len] : 9;
    ull ret = 0;
    for (int i=0; i<=Max; i++){
        ret += dfs(len-1, piovt, sum+(len-piovt)*i, limit&&(i==Max));
    }

    if (!limit){
        dp[len][piovt][sum] = ret;
    }
    return ret;
}

ull calc(ull n){

    if (n < 0){
        return 0;
    }

    int len = 0;

    while (n > 0){
        digit[++len] = n % 10;
        n /= 10;
    }

    ull ret = 0;
    for (int i=1; i<=len; i++){
        ret += dfs(len, i, 0, true);
    }

    return ret-(len-1);
}

int main()
{
    int t;
    ull n, m;

    scanf("%d", &t);
    memset(dp, -1, sizeof(dp));
    while (t > 0){
        t--;

        scanf("%I64d%I64d", &n,&m);
        printf("%I64d\n",calc(m)-calc(n-1));
    }
    return 0;
}