# An example of online pen test based on C++

• 2020-05-27 06:43:42
• OfStack

This paper illustrates the online pen test based on C++. I will share it with you for your reference as follows:

Recently in the Wolf factory internship, for a long time did not do the problem. Autumn recruit the first hair, pinduoduo... Four simple questions. Some people find it difficult? I will lower my RP by 1 shot. I think it is difficult for this topic. If you get a better Offer than me, I will not be convinced.

Four questions... Actually, I've only been writing for 40 minutes. But t in the end, there is no perfect score, 390/400, I don't know why the third question 1 straight 10 points can not pass.

One more. I think I just found number three... this > Number, I wrote > =... The & # 63; But they seem to me to be > = ah...

Question 1:

Time complexity O(n) and space complexity O(1) are required.

So there's actually two ways to do it, the three largest Numbers times |, | the two smallest Numbers times the largest number. Time complexity O(n), instantly thought of time complexity O(n) to find k large classical algorithm, divide and conquer!

``````
#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
const int N = 1e6 + 10;
long long a[N];
int k;
int partition(int l,int r) {
while(l != r)
{
while(a[r] >= a[l] && r > l)
r--;
if(l == r)
break;
swap(a[r],a[l]);
while(a[l] < a[r] && r > l)
l++;
if(l < r)
swap(a[r],a[l]);
}
return l;
}
long long solve(int l,int r) {
int now = partition(l,r);
if(k < now)
return solve(l,now-1);
else if(k > now)
return solve(now+1,r);
else
return a[now];
}
int main() {
int n;
while(~scanf("%d", &n)) {
for(int i = 0; i < n; ++i) {
scanf("%lld", &a[i]);
}
k = n - 1;
long long x1 = solve(0, n-1);
k = n - 2;
long long x2 = solve(0, n-2);
k = n - 3;
long long x3 = solve(0, n-3);
long long Ans = x1 * x2 * x3;
if(n > 3) {
k = 0;
long long y1 = solve(0, n-1);
k = 1;
long long y2 = solve(0, n-2);
Ans = max(Ans, y1*y2*x1);
}
printf("%lld\n", Ans);
}
return 0;
}

``````

Question 2:

Multiply large Numbers, template problem, found a template...

``````
#include <iostream>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int N = 1e5 + 10;
string c1, c2;
int a[N], b[N], r[N];
void solve(int a[], int b[], int la, int lb) {
int i, j;
for(i = 0; i != N; i++) r[i] = 0;
for(i = 0; i != la; i++)
{
for(j = 0; j != lb; j++)
{
int k = i + j;
r[k] += a[i] * b[j];
while(r[k] > 9)
{
r[k + 1] += r[k] / 10;
r[k] %= 10;
k++;
}
}
}
int l = la + lb - 1;
while(r[l] == 0 && l > 0) l--;
for(int i = l; i >= 0; i--) cout << r[i];
cout << endl;
}
int main() {
while(cin >> c1 >> c2)
{
int la = c1.size(), lb = c2.size();
for(int i = 0; i != la; i++)
a[i] = (int)(c1[la - i - 1] - '0');
for(int i = 0; i != lb; i++)
b[i] = (int)(c2[lb - i - 1] - '0');
solve(a, b, la, lb);
}
return 0;
}

``````

Question 3:

Greedy, I'm greedy by trying to meet the needs of the smallest people... 1 straight 90% & # 63; Some people try to use the biggest chocolate they can. 100%. How about a counter example?

``````
#include <cstdio>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
const int N = 3e6 + 10;
long long w[N], h[N];
int main() {
int n, m;
while(~scanf("%d", &n)) {
for(int i = 0; i < n; ++i) {
scanf("%lld", &h[i]);
}
scanf("%d", &m);
for(int i = 0; i < m; ++i) {
scanf("%lld", &w[i]);
}
sort(h, h + n);
sort(w, w + m);
int Ans = 0;
for(int i = 0, j = 0; i < n && j < m; ) {
if(w[j] >= h[i]) {
++Ans;
++i, ++j;
}
else {
++j;
}
}
printf("%d\n", Ans);
}
return 0;
}

``````

Question 4:

Maze problem. What's interesting is that there is an extra key... 1 janitor no more than 10... M, N < = 100... Think about the number of states... Direct state compression. Then there is a very violent and shameful state of compression bfs... Then I sent AC.

``````
#include <cstdio>
#include <iostream>
#include <string>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
const int N = 110;
char mz[N][N];
bool vis[N][N][N*10];
int fx = {0, 0, 1, -1};
int fy = {1, -1, 0, 0};
int m, n;
map<char, int> key;
struct node {
int x, y, cnt, sta;
node():cnt(0), sta(0) {}
};
queue<node> que;
int bfs(int sx, int sy, int ex, int ey) {
while(!que.empty()) que.pop();
node tmp;
tmp.x = sx, tmp.y = sy;
que.push(tmp);
while(!que.empty()) {
node p = que.front();
if(p.x == ex && p.y == ey) {
return p.cnt;
}
que.pop();
for(int i = 0; i < 4; ++i) {
int newx = p.x + fx[i];
int newy = p.y + fy[i];
if(newx < 0 || newx >= m || newy < 0 || newy >= n) continue;
if(mz[newx][newy] == '0') continue;
int sta = p.sta;
if(mz[p.x][p.y] >= 'a' && mz[p.x][p.y] <= 'z') {
sta |= (1<<key[mz[p.x][p.y]]);
}
if(vis[newx][newy][sta]) continue;
if(mz[newx][newy] >= 'A' && mz[newx][newy] <= 'Z') {
if((sta & (1<<(key[mz[newx][newy] - 'A' + 'a'])))== 0) {
continue;
}
}
vis[newx][newy][sta] = true;
tmp.x = newx, tmp.y = newy, tmp.cnt = p.cnt + 1, tmp.sta = sta;
que.push(tmp);
}
}
return -1;
}
int main() {
while(~scanf("%d %d", &m, &n)) {
int sx, sy, ex, ey;
int cnt = 0;
for(int i = 0; i < m; ++i) {
scanf("%s", mz[i]);
for(int j = 0; j < n; ++j) {
if(mz[i][j] == '2') {
sx = i, sy = j;
}
if(mz[i][j] == '3') {
ex = i, ey = j;
}
if(mz[i][j] >= 'a' && mz[i][j] <= 'z') {
key[mz[i][j]] = cnt++;
}
}
}
for(int i = 0; i < m; ++i) {
for(int j = 0; j < n; ++j) {
for(int s = 0; s < (1<<cnt); ++s) {
vis[i][j][s] = false;
}
}
}
int Ans = bfs(sx, sy, ex, ey);
printf("%d\n", Ans);
}
return 0;
}

``````