[Bzoj 1833] [ZJOI2010] Count number count "digital DP"

Title Link: BZOJ-1833

Problem analysis

Digital DP..

Using F[i][j][k] Indicates the number of digits K for the i digit of J.

Then differential inquiry ... Get () Note that if the bit I is fixed, this is T, then the answer to the digital T is to add a value (see Code).

Code

#include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <cmath > #include <algorithm>using namespace std;const int maxbit = 15;typedef long long ll;struct es{ll a[11];}; ES operator + (const es &a, const es &b) {ES ret;for (int i = 0; I <= 9; ++i) ret. A[i] = A.a[i] + b.a[i];return ret;} LL A, B; LL P10[maxbit]; Es f[maxbit][11];es Get (LL x) {ES ret;for (int i = 0; I <= 9; ++i) ret. A[i] = 0;if (x = = 0) return ret;int L = 1;while (P10[l] <= x) ++l;for (int i = 1; I <= l-1; ++i) {for (int j = 1; J <= 9; ++J) {ret = ret + f[i][j];}} 0 were not counted ++ret. A[0]; LL T; t = x/p10[l-1];x%= p10[l-1];//If there is only 1 bits, the following will not be counted 0, but already on the above 0 for (int i = 1; I <= t-1; ++i) ret = ret + f[l][i] ; ret. A[t] + = x;for (int i = L-1; I >= 1; i.) {t = x/p10[i-1];x%= p10[i-1];for (int j = 0; J <= t-1; ++j) ret = ret + F[i][j];ret. A[T] + = x;} return ret;} int main () {p10[0] = 1LL; for (int i = 1; i <=; ++i) P10[i] = P10[I-1] * 10ll;for (int i = 1; i <=; ++i) {for (int j = 0; J <= 9; ++j) {for (int k = 0; k <= 9; ++k) {F[i ][J] = F[i][j] + f[i-1][k];} F[I][J]. A[J] + = P10[i-1];}} scanf ("%lld%lld", &a, &b); ES TA, Tb;ta = Get (A); TB = Get (B + 1), for (int i = 0; I <= 9; ++i) {printf ("%lld", TB. A[i]-TA. A[i]); if (i = = 9) printf ("\ n"), Else printf ("");} return 0;}

　　

[Bzoj 1833] [ZJOI2010] Count number count "digital DP"