SPOJ Balanced Numbers(数位dp,三进制状压)
2016-09-02 17:31
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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[L,R],求Balanced Number的数量。数据范围:1≤L≤R≤1019
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i位如果是0那就是代表出现偶数次,如果是1那就代表出现奇数次,然后滚动下就好了。但是0的情况不好判断是否出现,而且记忆化的时候也会出问题。
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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