[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 }

View Code

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！