折半搜索 (meet in the middle)
/*
reference:
translation:
solution:
trigger:
note:
*
date:
2019.09.04
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define dwn(i,a,b) for(int i=a;i>=b;--i)
template <typename T> inline void rd(T &x){x=0;char c=getchar();int f=0;while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}x=f?-x:x;}
#define mem(a,b) memset(a,b,sizeof(a))
#define ee(i,x) for(int i=head[x];i;i=e[i].next)
int n,t;
int sum,tmp;
int a[50];
bool found=false;
map<int,bool> mp;
inline void dfs1(int h,int t){
if(h>t){
mp[tmp]=1;
return ;
}
dfs1(h+1,t);
tmp+=a[h];
dfs1(h+1,t);
tmp-=a[h];
}
inline void dfs2(int h,int t){
if(h>t){
if(mp[sum-tmp])found=1;
return ;
}
dfs2(h+1,t);
tmp+=a[h];
dfs2(h+1,t);
tmp-=a[h];
}
#undef int
int main(){
#define int long long
rd(n),rd(sum);
rep(i,1,n)rd(a[i]);
dfs1(1,n/2);
dfs2(n/2+1,n);
printf(found?"YES":"NO");
return 0;
}
/*
5 67
34 546 5 35 32
*/
CF888E Maximum_Subsequence
/*
reference:
translation:
solution:
考虑到dfs的效率很低很低而且mod数在1e9的范围,肯定要用一个stl的容器啊(set)
2的35次方会超时,考虑折半搜索,前后分别枚举,最后二分取最大值即可。
trigger:
note:
*
date:
2019.09.04
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define dwn(i,a,b) for(int i=a;i>=b;--i)
template <typename T> inline void rd(T &x){x=0;char c=getchar();int f=0;while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}x=f?-x:x;}
#define mem(a,b) memset(a,b,sizeof(a))
#define ee(i,x) for(int i=head[x];i;i=e[i].next)
const int N = 40;
int a[N],n,mod,ans;
set<int>s,t;
inline void dfs1(int cur,int sum){
if(cur==n>>1|1){
sum=sum%mod;
s.insert(sum);
return ;
}
dfs1(cur+1,sum);
dfs1(cur+1,sum+a[cur]);
}
inline void dfs2(int cur,int sum){
if(cur==n+1){
sum=sum%mod;
t.insert(sum);
return ;
}
dfs2(cur+1,sum);
dfs2(cur+1,sum+a[cur]);
}
#undef int
int main(){
#define int long long
rd(n),rd(mod);
rep(i,1,n){
rd(a[i]);
}
dfs1(1,0);
dfs2(n>>1|1,0);
for(auto it=s.begin();it!=s.end();++it){
int tmp=*it;
auto pos=t.lower_bound(mod-tmp);
if(pos!=t.end()){
--pos;
if(*pos+tmp<=mod)
ans=max(ans,*pos+tmp);
}
}
printf("%lld",ans);
return 0;
}
法2:two_pointers
/*
4 4
5 2 4 1
*/
//3
/*
3 20
199 41 299
*/
//19
/*
reference:
translation:
solution:
法2:two_pointer 来找两个区间的在一定值的限定区间的最大值,如本题要求
在x数组和y数组(要排个序)中各选择一个数的和<=mod-1,并使这个值最大
很显然,快指针从前往后,慢指针倒序,如果当前值比mod-1大了那么j--,因为i再怎么往后移值都不可能
<=mod-1,(因为是按照升序排序的)
trigger:
note:
*注意最大值<=mod-1而不是<=mod,你太瓜啦
date:
2019.09.04
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define dwn(i,a,b) for(int i=a;i>=b;--i)
template <typename T> inline void rd(T &x){x=0;char c=getchar();int f=0;while(!isdigit(c)){f|=c=='-';c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}x=f?-x:x;}
#define mem(a,b) memset(a,b,sizeof(a))
#define ee(i,x) for(int i=head[x];i;i=e[i].next)
const int N = 40;
int a[N],x[N],y[N],n,mod,ans;
int tot1,tot2;
//set<int>s,t;
inline void dfs1(int cur,int sum){
if(cur==n/2+1){
sum=sum%mod;
x[++tot1]=sum;
//s.insert(sum);
return ;
}
dfs1(cur+1,sum);
dfs1(cur+1,sum+a[cur]);
}
inline void dfs2(int cur,int sum){
if(cur==n+1){
sum=sum%mod;
y[++tot2]=sum;
//t.insert(sum);
return ;
}
dfs2(cur+1,sum);
dfs2(cur+1,sum+a[cur]);
}
#undef int
int main(){
#define int long long
freopen("cf888e.txt","r",stdin);
rd(n),rd(mod);
rep(i,1,n){
rd(a[i]);
}
dfs1(1,0);
dfs2(n>>1|1,0);
/*for(auto it=s.begin();it!=s.end();++it){
int tmp=*it;
auto pos=t.lower_bound(mod-tmp);
if(pos!=t.end()){
--pos;
if(*pos+tmp<=mod)/////////////////这里应该是<
ans=max(ans,*pos+tmp);
}
}*/
sort(x+1,x+tot1+1);
sort(y+1,y+tot2+1);
/*rep(i,1,tot1)printf("%lld ",x[i]);
puts("");
rep(i,1,tot2)printf("%lld ",y[i]);
puts("");*/
for(int i=1,j=tot2;i<=tot1;++i){
if(x[i]+y[j]>=ans && x[i]+y[j]<=mod-1)
ans=x[i]+y[j];
while(j && x[i]+y[j]<ans)
j--;
}
printf("%lld",ans);
return 0;
}
原文地址:https://www.cnblogs.com/sjsjsj-minus-Si/p/11634700.html
时间: 2024-11-05 23:22:34