Spoj myq10 10649. Mirror number (Digital DP)

Spoj myq10 10649. Mirror number (Digital DP)

ACM

Address: spoj myq10 mirror number

Question:
Calculate the number of image replies in [a, B.
0 <= A <= B <= 10 ^ 44

Analysis:
When you see the question and data range, you will know that it is a digital DP.
Obviously, the image only returns 0, 1, and 8. Just like the reply question, you have to set up an auxiliary array to record the number selected in front of the DFS.
Note that the leading 0 must be removed.

Code:

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        10649_MYQ10.cpp*  Create Date: 2014-08-02 14:20:31*  Descripton:  MYQ10, digit dp*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define repf(i,a,b) for(int i=(a);i<=(b);i++)typedef long long ll;const int N = 50;ll dp[N][N][2];// dp[start][current][ismirror]char bit[N], num[N];int t, rec[N], len;// the start pos, current pos, ismirror flag, is limit flag// this function will use rec[] to record the prevent chosen numbersll dfs (int start, int cur, bool ismir, bool limit) {if (cur < 0)return ismir;if (!limit && dp[start][cur][ismir] != -1)return dp[start][cur][ismir];int mmax = limit ? bit[cur] - '0' : 9;ll ret = 0;repf (i, 0, mmax) {if (i == 0 || i == 1 || i == 8) {rec[cur] = i;if (start == cur && i == 0) {// if have lead zeroret += dfs(start - 1, cur - 1, ismir, (limit && (i == mmax)));} else if (ismir && cur < (start + 1) / 2) {// if ismirror and can judgeret += dfs(start, cur - 1, (i == rec[start - cur]), (limit && (i == mmax)));} else {ret += dfs(start, cur - 1, ismir, (limit && (i == mmax)));}}}return limit ? ret : dp[start][cur][ismir] = ret;}// check if first number is mirrorbool check() {repf (i, 0, len >> 1) {if (bit[i] != '0' && bit[i] != '1' && bit[i] != '8')return false;if  (bit[i] != bit[len - 1 - i])return false;}return true;}ll solve() {ll ans1, ans2;// first numberscanf("%s", num);len = strlen(num);repf (i, 0, len - 1)bit[i] = num[len - 1 - i];bit[len] = 0;ans1 = dfs(len - 1, len - 1, 1, 1) - check();//cout << ans1 << endl;// second numberscanf("%s", num);len = strlen(num);repf (i, 0, len - 1)bit[i] = num[len - 1 - i];bit[len] = 0;ans2 = dfs(len - 1, len - 1, 1, 1);//cout << ans2 << endl;cout << ans2 - ans1 << endl;}int main() {memset(dp, -1, sizeof(dp));scanf("%d", &t);while (t--) {solve();}return 0;}