[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。

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21
 22 using namespace std;
 23
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define mem_1(a) memset(a, -1, sizeof(a))
 26 #define lson l, m, rt << 1
 27 #define rson m + 1, r, rt << 1 | 1
 28 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 29 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 30 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 31 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 32 #define all(a) (a).begin(), (a).end()
 33 #define lowbit(x) ((x) & (-(x)))
 34 #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) {}
 35 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 36 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 37 #define pchr(a) putchar(a)
 38 #define pstr(a) printf("%s", a)
 39 #define sstr(a) scanf("%s", a)
 40 #define sint(a) scanf("%d", &a)
 41 #define sint2(a, b) scanf("%d%d", &a, &b)
 42 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 43 #define pint(a) printf("%d\n", a)
 44 #define test_print1(a) cout << "var1 = " << a << endl
 45 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 46 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 47 #define mp(a, b) make_pair(a, b)
 48 #define pb(a) push_back(a)
 49
 50 typedef unsigned int uint;
 51 typedef long long LL;
 52 typedef pair<int, int> pii;
 53 typedef vector<int> vi;
 54
 55 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 56 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 57 const int maxn = 1e5 + 7;
 58 const int md = 100000007;
 59 const int inf = 1e9 + 7;
 60 const LL inf_L = (LL)1e18 + 7;
 61 const double pi = acos(-1.0);
 62 const double eps = 1e-6;
 63
 64 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 65 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 66 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 67 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 68 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 69 int make_id(int x, int y, int n) { return x * n + y; }
 70
 71 struct Node {
 72     LL a[808];
 73     LL &operator [] (const int x) {
 74         return a[x + 401];
 75     }
 76 } dp[20][20];
 77
 78 void div_digit(LL x, int a[], int &n) {
 79     int p = 0;
 80     a[p ++] = x % 10;
 81     x /= 10;
 82     while (x) {
 83         a[p ++] = x % 10;
 84         x /= 10;
 85     }
 86     n = p;
 87 }
 88
 89 void init() {
 90     rep_up0(p, 18) dp[p][0][0] = 1;
 91     rep_up0(p, 18) {
 92         rep_up1(i, 17) {
 93             for (int j = -400; j <= 400; j ++) {
 94                 rep_up0(k, 10) {
 95                     int val = j + (p - i + 1) * k;
 96                     if (val < -400 || val > 400) continue;
 97                     dp[p][i][j] += dp[p][i - 1][val];
 98                 }
 99             }
100         }
101     }
102 }
103
104 LL calc(LL n) {
105     if (n == -1) return 0;
106     if (n == (LL)1e18) return (LL)12644920956811384;
107     LL ans = 0;
108     int a[20], len;
109     div_digit(n, a, len);
110     rep_up0(p, 18) {
111         int sum = 0;
112         rep_down0(i, len) {
113             rep_up0(j, a[i]) {
114                 int val = sum + (p - i) * j;
115                 if (val < -400 || val > 400) continue;
116                 ans += dp[p][i][val];
117             }
118             sum += a[i] * (p - i);
119         }
120     }
121     rep_up0(p, 18) {
122         int sum = 0;
123         rep_down0(i, len) {
124             sum += a[i] * (p - i);
125         }
126         if (sum == 0) ans ++;
127     }
128     return ans - 17;
129 }
130
131 int main() {
132     //freopen("in.txt", "r", stdin);
133     init();
134     int T;
135     cin >> T;
136     LL n, m;
137     while (T --) {
138         cin >> n >> m;
139         cout << calc(m) - calc(n - 1) << endl;
140     }
141     return 0;
142 }

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！