SPOJ Balanced Numbers(数位dp,三进制状压)
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题目链接
SPOJ Balanced Numbers
题意
定义一个数如果满足所有位上偶数数字出现奇数次,奇数数字出现偶数次那么就称为Balanced Number。给一个区间
数据范围:
分析
根据数据范围,推荐使用unsigned long long
+cin/cout
。
满足区间减法。仍然是状压的思路,一开始我是想状压成二进制数:二进制数的第
改成三进制就好了。0代表没出现,1代表出现奇数次,2代表出现偶数次。注意前导0。
Code
#include <stdio.h>#include <string.h>#include <algorithm>#include <math.h>#include <climits>#include <iostream>using namespace std;typedef unsigned long long ll;int T, digit[25], cnt[15], pw3[15];ll A, B, dp[20][60000];int check(int state){ for (int i = 0; i < 10; ++i) { int t = state % 3; if ((i % 2 == 1) && t == 1) return 0; if ((i % 2 == 0) && t == 2) return 0; state /= 3; } return 1;}ll dfs(int pos, int state, int first, int limit){ if (pos == -1) return check(state); if (!limit && dp[pos][state] != -1) return dp[pos][state]; int last = limit ? digit[pos] : 9; ll ret = 0; for (int i = 0; i <= last; ++i) { int cur = state / pw3[i] % 3, nxt = state; if (!(i == 0 && first)) { if (cur == 0 || cur == 1) nxt += pw3[i]; else nxt -= pw3[i]; } ret += dfs(pos - 1, nxt, first && (i == 0), limit && (i == last)); } if (!limit) dp[pos][state] = ret; return ret;}ll solve (ll x){ memset(digit, 0, sizeof (digit)); int len = 0; while (x) { digit[len++] = x % 10; x /= 10; } return dfs(len - 1, 0, 1, 1);}int main(){ ios_base::sync_with_stdio(0); cin.tie(0); pw3[0] = 1; for (int i = 1; i <= 10; ++i) { pw3[i] = 3 * pw3[i - 1]; } cin >> T; memset(dp, -1, sizeof (dp)); while (T--) { cin >> A >> B; cout << solve(B) - solve(A - 1) << endl; } return 0;}
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