P4310 绝世好题 按位DP

这名字可海星\(OvO\)

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思路：\(DP\)

提交：2次（\(zz\)我竟然把三目运算符写错了\(QwQ\)）

题解：

按位进行\(DP\)：\(f[i]\)表示结尾的数字包括\(1<<i\)的最长合法序列的长度。所以每输进来一个数\(x\)，我们就可以尝试更新\(f[i]=max(\)所有的 \(f[i])\) 且 \(x\ and\ (1<<i)!=0\)。

没了。

代码

#include<cstdio>
#include<iostream>
#define ull unsigned long long
#define ll long long
#define R register int
using namespace std;
#define pause (for(R i=1;i<=10000000000;++i))
#define In freopen("NOIPAK++.in","r",stdin)
#define Out freopen("out.out","w",stdout)
namespace Fread {
static char B[1<<15],*S=B,*D=B;
#ifndef JACK
#define getchar() (S==D&&(D=(S=B)+fread(B,1,1<<15,stdin),S==D)?EOF:*S++)
#endif
inline int g() {
	R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
	if(ch==EOF) return EOF; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
} inline bool isempty(const char& ch) {return (ch<=36||ch>=127);}
inline void gs(char* s) {
	register char ch; while(isempty(ch=getchar()));
	do *s++=ch; while(!isempty(ch=getchar()));
}
} using Fread::g; using Fread::gs;

namespace Luitaryi {
int n,f[32],ans;
inline void main() {
	n=g(); for(R i=1,x;i<=n;++i) { R tmp=0;
		x=g(); for(R i=0;i<=30;++i) tmp=max(tmp,((1<<i)&x?1:0)*f[i]+1);
		for(R i=0;i<=30;++i) f[i]=max(((1<<i)&x?1:0)*tmp,f[i]);
	} for(R i=0;i<=30;++i) ans=max(ans,f[i]); printf("%d\n",ans);
}
}
signed main() {
	Luitaryi::main();
	return 0;
}

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2019.07.22