Single Round Match 508 Round 1 - Division I, Level Two YetAnotherORProblem
2014-11-26 19:09
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很容易发现,满足题目性质就意味着所有位的加法不能产生进位。于是所有位只有两种情况,都是0,或者是只有一个1。我没可以采用类似数位DP的方法,从高位向低位考虑。dp[i][j]表示考虑前i个高位,j是一个集合,表示n个数字的状态。如果第k的数字的前i的高位已经小于原数了就是1,反之是0。因为在转移的时候,j位上是1的一定可以在这一位上放1,否则当且仅当原数的第i位也是1才能放1。转移分两种情况,一个是全部放0,一个是放一个1,注意集合j的转移就可以了。
#include <bits/stdc++.h> using namespace std; long long dp[70][1<<10]; const int MOD=1000000009; class YetAnotherORProblem { public: int countSequences(vector<long long> a) { int n=a.size(); dp[63][0]=1; for(int i=63;i>0;i--) { for(int j=0;j<(1<<n);j++) { if(!dp[i][j]) continue; int cur=j; for(int k=0;k<n;k++) { if(a[k]&(1LL<<(i-1))) cur|=(1<<k); } dp[i-1][cur]=(dp[i-1][cur]+dp[i][j])%MOD; for(int k=0;k<n;k++) { if((1<<k)&j) { int st=j; for(int l=0;l<n;l++) { if(l==k) continue; if(a[l]&(1LL<<(i-1))) { st|=(1<<l); } } dp[i-1][st]=(dp[i-1][st]+dp[i][j])%MOD; } else { if(a[k]&(1LL<<(i-1))) { int st=j; for(int l=0;l<n;l++) { if(l==k) continue; if(a[l]&(1LL<<(i-1))) { st|=(1<<l); } } dp[i-1][st]=(dp[i-1][st]+dp[i][j])%MOD; } } } } } long long ans=0; for(int i=0;i<1<<n;i++) ans=(ans+dp[0][i])%MOD; return ans; } };
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