寒武纪 2019.9.17笔试 3.2
5选择4编程
选择很水
编程题有会的请告诉我。。。
第一题，
给AB两个队和一个target，把数字往队里放，如果每个队对每个数，都有另一个数对应它门的和是target，那么这个队好。两个队对都好还是啥的，忘了。
按贪心随便写了些，懒得改交卷了全局情况0.3
#include<iostream>
#include<string>
#include<vector>
#include<map>
using namespace std;
void res(int N,int X,int Y,vector<int>&input)
{
int flag = false;
map<int,int>record;
for(int i=0;i<N;++i)
record[input[i]]=-1;
//record.insert({input[i],-1});
//int index=0;
for(int i=0;i<N;++i)
{
if(record[input[i]]!=-1)
continue;
if(record.find(X-input[i])!=record.end())
{
if(record[X-input[i]]==-1)
{
record[X-input[i]]=X;
record[input[i]]=X;
}
else if(record[X-input[i]]==Y)
{
//index=i;
flag=true;
break;
}
}else if(record.find(Y-input[i])!=record.end())
{
if(record[Y-input[i]]==-1)
{
record[Y-input[i]]=Y;
record[input[i]]=Y;
}
else if(record[Y-input[i]]==X)
{
flag=true;
//index=i;
break;
}
}else
{
flag = true;
break;
}
}
if(flag)
{
cout<<"NO"<<endl;
//cout<<index<<endl;
//cout<<record[index]<<endl;
return ;
}
cout<<"YES"<<endl;
for(int i=0;i<N-1;++i)
{
cout<<record[input[i]]<<" ";
}
cout<<record[input[N-1]]<<endl;
return;
}
int main()
{
int N,X,Y;
cin>>N>>X>>Y;
vector<int>data(N,0);
for(int i=0;i<N;++i)
{
cin>>data[i];
}
res(N,X,Y,data);
}
第二题 给一些数，然后压缩成什么东西，可以选一个上下阈值，两边删掉，问最少删多少个
滑窗，可以改成二分0.9
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<algorithm>
typedef long long LL;
using namespace std;
LL maxmax(LL &a,LL &b)
{
if(a<b)return b;
return a;
}
void print_arr(vector<LL>&data)
{
for(auto c:data)cout<<c<<" ";
cout<<endl;
}
LL res(vector<LL>input,LL N,LL M)
{
M*=8;
LL bitNum=M/N;
LL nums;
if(bitNum==0)
{
//cout<<'a'<<endl;
return N-1;
}
if(bitNum<31)
nums = pow(2,bitNum);
else
{
//cout<<'b'<<endl;
return 0;
}
sort(input.begin(),input.end());
vector<LL>arrNum(1,0);
vector<LL>arrGeshu(1,0);
for(int i=0;i<N;)
{
int j=i;
while(j<N && input[i]==input[j])++j;
arrNum.push_back(input[i]);
arrGeshu.push_back(j);
i=j;
}
//print_arr(input);
//print_arr(arrGeshu);
//无符号比LL,害怕
if(arrNum.size()<nums)
{
//cout<<'c'<<endl;
return 0;
}
LL t_max=0;
for(unsigned int i=0;i<arrNum.size()-nums;++i)
{
//if(arrGeshu[i+nums]-arrGeshu[i]>0)cout<<'>'<<endl;
if(t_max<arrGeshu[i+nums]-arrGeshu[i])t_max=arrGeshu[i+nums]-arrGeshu[i];
}
//t_max = maxmax(t_max,arrGeshu[i+nums]-arrGeshu[i]);
return N-t_max;
}
int main()
{
LL N,M;
cin>>N>>M;
vector<LL>data(N,0);
for(int i =0;i<N;++i)cin>>data[i];
cout<<res(data,N,M)<<endl;
}
字符串匹配，B串在A串顺序出现 1.0
#include<iostream>
#include<string>
using namespace std;
string res(string &s1,string &s2)
{
if(s2.size()>s1.size())
{
return "NO";
}
int cur = 0;
int len = s2.size();
for(auto c:s1)
{
if(cur==len)
return "SUB";
if(c==s2[cur])
++cur;
}
if(cur==len)
return "SUB";
return "NO";
}
int main()
{
int n;
cin>>n;
while(n--)
{
string s1,s2;
cin>>s1>>s2;
cout<<res(s1,s2)<<endl;
} }
池化层手撸1.0
#include<iostream>
#include<string>
#include<vector>
using namespace std;
void print_res(vector<vector<int>>&data,int hang,int lie)
{
for(int i=0;i<hang;i++)
{
for(int j=0;j<lie-1;j++)
{
cout<<data[i][j]<<" ";
}
cout<<data[i][lie-1]<<endl;
}
}
int main()
{
int input_h,input_l;
cin>>input_h>>input_l;
int chi_h,chi_l;
cin>>chi_h>>chi_l;
int stride_h,stride_l;
cin>>stride_h>>stride_l;
vector<vector<int>>data;
for(int i=0;i<input_h;++i)
{
vector<int>tmp(input_l,0);
for(int j=0;j<input_l;++j)
{
cin>>tmp[j];
}
data.push_back(tmp);
}
int output_h = 1+(input_h-chi_h)/stride_h;
int output_l = 1+(input_l-chi_l)/stride_l;
vector<vector<int>>out(output_h,vector<int>(output_l,0));
for(int i=0;i<output_h;++i)
{
for(int j=0;j<output_l;++j)
{
int start_h = i*stride_h;
int start_l = j*stride_l;
int res = -0x3f3f3f3f;
for(int ii=start_h;ii<start_h+chi_h;++ii)
for(int jj=start_l;jj<start_l+chi_l;++jj)
{
res = max(res,data[ii][jj]);
}
out[i][j]=res;
}
}
print_res(out,output_h,output_l);
}
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