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SPOJ BALNUM Balanced Numbers 状压+数位DP
一开始想了一个用二进制状压的方法,发现空间需要的太大,光光memset都要超时 = = 其实不用每次都memset
也可以用三进制,一开始直接打表出所有的状态转移就好
#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <climits>#include <string>#include <iostream>#include <map>#include <cstdlib>#include <list>#include <set>#include <queue>#include <stack>using namespace std;typedef unsigned long long ULL;typedef long long LL;const ULL INF = ~0ULL;const int maxn = 20;const int maxstate = 59049;int lim[maxn],len = 0;int ccon[maxstate][10];bool ok[maxstate];void getlim(ULL num) { memset(lim,0,sizeof(lim)); len = 0; while(num) { lim[len++] = num % 10; num /= 10; }}ULL f[maxn][maxstate];int tmp[10];inline int con(int ns,int pos) { for(int i = 9;i >= 0;i--) { tmp[i] = ns % 3; ns /= 3; } if(tmp[pos] == 0) tmp[pos] = 1; else if(tmp[pos] == 1) tmp[pos] = 2; else tmp[pos] = 1; int ret = 0; for(int i = 0;i <= 9;i++) { ret *= 3; ret += tmp[i]; } return ret;}inline bool check(int st) { for(int i = 9;i >= 0;i--) { tmp[i] = st % 3; st /= 3; } for(int i = 0;i <= 9;i++) if(tmp[i]) { if((i % 2) == (tmp[i] % 2)) return false; } return true;}ULL dfs(int now,int state,int first,int bound) { if(now == 0) { return ok[state]; } ULL ¬e = f[now][state]; if(!bound && first && note != INF) return note; int m = bound ? lim[now - 1] : 9; ULL ret = 0; for(int i = 0;i <= m;i++) { if(first || i) { ret += dfs(now - 1,ccon[state][i],1,i == m && bound); } else { ret += dfs(now - 1,state,0,i == m && bound); } } if(!bound && first) note = ret; return ret;}ULL solve(ULL num) { getlim(num); return dfs(len,0,0,1);}void init() { for(int i = 0;i < maxstate;i++) { for(int j = 0;j < 10;j++) { ccon[i][j] = con(i,j); } ok[i] = check(i); }}int main() { int T; cin >> T; ULL a,b; init(); memset(f,0xff,sizeof(f)); while(T--) { cin >> a >> b; cout << solve(b) - solve(a - 1) << endl; } return 0;}
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