POJ 2282 数位dp

#include <cstdio>#include <cstring>#include <algorithm>const int maxn = 12;int A[maxn], B[maxn], a, b;void Fun(int *L, int n, int m){int x = n / 10, y = n % 10, tmp = x;for (int i = 0; i <= y; i++) L[i] += m;for (int i = 0; i < 10; i++) L[i] += m * x;L[0] -= m;while (tmp) L[tmp % 10] += m * (y + 1), tmp /= 10;if (x) Fun(L, x - 1, m * 10);}int main(int argc, char const *argv[]){while (~scanf("%d%d", &a, &b) && a + b){if (a > b) std::swap(a, b);memset(A, 0, sizeof(A));memset(B, 0, sizeof(B));Fun(A, a - 1, 1); Fun(B, b, 1);for (int i = 0; i < 10; i++)printf("%d%c", B[i] - A[i], i == 9 ? '\n' : ' ');}return 0;}

f(1234) 可以从1231,1232,1233,1234，数四个，然后可以算f(1230) ，f(1230) 可以由f(123) 得到。注意去除前导0。