LightOJ 1205 - Palindromic Numbers （数位dp）

ACM

题目地址：SPOJ MYQ10 Mirror Number

题意： 
求[a,b]中回文的个数。

分析： 
是SPOJ MYQ01的简单版...其实有非递归方法的。

代码：

/**  Author:      illuz <iilluzen[at]gmail.com>*  Blog:        http://blog.csdn.net/hcbbt*  File:        1205.cpp*  Create Date: 2014-08-02 16:21:04*  Descripton:  digit dp, palindrome */#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 = 20;int t, len, rec[N], bit[N];ll a, b, dp[N][N][2];ll 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] : 9;ll ret = 0;repf (i, 0, mmax) {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;}ll calc(ll n) {len = 0;while (n) {bit[len++] = n % 10;n /= 10;}bit[len] = 0;return dfs(len - 1, len - 1, 1, 1);}bool check() {repf (i, 0, len / 2 - 1) {if (bit[i] != bit[len - 1 - i])return false;}return true;}int main() {memset(dp, -1, sizeof(dp));scanf("%d", &t);int cas = 1;while (t--) {scanf("%lld %lld", &a, &b);if (a < b) swap(a, b);printf("Case %d: %lld\n", cas++, calc(a) - calc(b) + check());}return 0;}