ACM
题意:
求[a,b]中镜像回文的个数。
0 <= a<=b <= 10^44
分析:
看到题目和数据范围就知道是数位dp了。
很明显镜像回文只有0,1,8,跟回文的一题一样,在dfs的时候得开个辅助数组记录前面已经选择的数字。
注意还得去掉前导0。
代码:
/* * 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 numbers 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] - '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 zero ret += dfs(start - 1, cur - 1, ismir, (limit && (i == mmax))); } else if (ismir && cur < (start + 1) / 2) { // if ismirror and can judge ret += 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 mirror bool 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 number scanf("%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 number scanf("%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; }
SPOJ MYQ10 10649. Mirror Number (数位dp),布布扣,bubuko.com
SPOJ MYQ10 10649. Mirror Number (数位dp)
原文:http://blog.csdn.net/hcbbt/article/details/38349367