[zoj 3416/hdu 3709]数位DP

题意：求从区间[L, R]内有多少个数是平衡数，平衡数是指以10进制的某一位为中心轴，左右两边的每一位到中心轴的距离乘上数位上的值的和相等。0<=L<=R<=1e18

思路：由于任何非0正数最多只有1个位置作为中心轴使得它是平衡数。于是可以按中心轴的位置分类统计答案。令dp[p][i][j]表示中心轴在p位（p>=0)前i位且左边比右边的加权和已经多j的方案数，枚举当前第i位放的数k，那么dp[p][i][j]=∑dp[p][i-1][j+(p-i+1)*k]。

求出dp值后，只需从高位向低位统计，统计时也是按中心轴分类，即枚举中心轴，然后根据前多少位相同，维护一下已确定的数到中心轴的加权和，然后加上对应dp值。由于0这个数无论以什么作为中心轴都会使答案加1，所以最后需减去重复的。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d\n", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)

typedef unsigned int uint;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 1e5 + 7;
const int md = 100000007;
const int inf = 1e9 + 7;
const LL inf_L = (LL)1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

struct Node {
    LL a[808];
    LL &operator [] (const int x) {
        return a[x + 401];
    }
} dp[20][20];

void div_digit(LL x, int a[], int &n) {
    int p = 0;
    a[p ++] = x % 10;
    x /= 10;
    while (x) {
        a[p ++] = x % 10;
        x /= 10;
    }
    n = p;
}

void init() {
    rep_up0(p, 18) dp[p][0][0] = 1;
    rep_up0(p, 18) {
        rep_up1(i, 17) {
            for (int j = -400; j <= 400; j ++) {
                rep_up0(k, 10) {
                    int val = j + (p - i + 1) * k;
                    if (val < -400 || val > 400) continue;
                    dp[p][i][j] += dp[p][i - 1][val];
                }
            }
        }
    }
}

LL calc(LL n) {
    if (n == -1) return 0;
    if (n == (LL)1e18) return (LL)12644920956811384;
    LL ans = 0;
    int a[20], len;
    div_digit(n, a, len);
    rep_up0(p, 18) {
        int sum = 0;
        rep_down0(i, len) {
            rep_up0(j, a[i]) {
                int val = sum + (p - i) * j;
                if (val < -400 || val > 400) continue;
                ans += dp[p][i][val];
            }
            sum += a[i] * (p - i);
        }
    }
    rep_up0(p, 18) {
        int sum = 0;
        rep_down0(i, len) {
            sum += a[i] * (p - i);
        }
        if (sum == 0) ans ++;
    }
    return ans - 17;
}

int main() {
    //freopen("in.txt", "r", stdin);
    init();
    int T;
    cin >> T;
    LL n, m;
    while (T --) {
        cin >> n >> m;
        cout << calc(m) - calc(n - 1) << endl;
    }
    return 0;
}

另外，灵机一动想出来一种写法（如有雷同，纯属巧合），用起来也还不错哦！