HDU 3709 Balanced Number (数位DP)

题意：找出区间内平衡数的个数，所谓的平衡数，就是以这个数字的某一位为支点，另外两边的数字大小乘以力矩之和相等，即为平衡数。

析：数位DP，dp[i][[j][k]表示 前 i 位以 j 为支点，还差 k 平衡，枚举 j 就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[20][20][1500];
int a[20];

LL dfs(int pos, int dot, int balance, bool ok){
    if(!pos)  return balance == 0;
    if(balance < 0)  return 0;
    LL &ans = dp[pos][dot][balance];
    if(!ok && ans >= 0)  return ans;

    LL res = 0;
    int n = ok ? a[pos] : 9;
    for(int i = 0; i <= n; ++i){
        res += dfs(pos-1, dot, balance+(pos-dot)*i, ok && i == n);
    }
    if(!ok)  ans = res;
    return res;
}

LL solve(LL n){
    int len = 0;
    while(n){
        a[++len] = n % 10;
        n /= 10;
    }
    LL ans = 0;
    for(int i = 1; i <= len; ++i)
        ans += dfs(len, i, 0, true);

    return ans - len;//除去全是0的情况
}

int main(){
    memset(dp, -1, sizeof dp);
    LL x, y;
    int T;  cin >> T;
    while(T--){
        scanf("%lld %lld", &x, &y);
        printf("%lld\n", solve(y)-solve(x-1));
    }
    return 0;
}