[置顶] 今日头条2018春季校园招聘研发岗位笔试 题解 临时版

前言

算算总分，98分，很难受。。。
临时版的题解都是只会把我的代码发上去，有问题的童鞋们可以留言，我尽力回答=w=

第一题

。。。这题很诡异，我只过了80%的数据，用set写会TLE，用hash写会WA，而且都是80%。。。

TLE版

#include <bits/stdc++.h>
using namespace std;
set<int> st;
set<pair<int, int> > ans;
int main() {
int n, k, x;
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++) {
scanf("%d", &x);
if (st.find(x + k) != st.end()) ans.insert(make_pair(x + k, x));
if (st.find(x - k) != st.end()) ans.insert(make_pair(x - k, x));
st.insert(x);
}
printf("%d\n", ans.size());
return 0;
}

WA版

#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e8 + 5;
const int INF = 1e8;
bool vis[MAX_N];
set<int> st[2];
int main() {
int n, k, x;
scanf("%d%d", &n, &k);
for (int i = 0; i < n; i++) {
scanf("%d", &x);
if (x - k >= 0 && vis[x - k]) st[0].insert(x);
if (k != 0 && x + k <= INF && vis[x + k]) st[1].insert(x);
vis[x] = true;
}
printf("%d\n", st[0].size() + st[1].size());
return 0;
}

第二题

简单的广搜就可以，为了保险我加了一个记忆化，不知道不加可不可以。#include <bits/stdc++.h>
using namespace std;
#define pr make_pair
#define fi first
#define se second
typedef pair<int, int> pii;
set<pii> st;
queue<pair<pii, int> > que;
int main() {
int n, s, m, v;
scanf("%d", &n);
que.push(pr(pr(1, 1), 0));
st.insert(pr(1, 1));
while (!que.empty()) {
s = que.front().fi.fi;
m = que.front().fi.se;
v = que.front().se;
//printf("s = %d, m = %d, v = %d\n", s, m, v);
que.pop();
if (s == n) {
printf("%d\n", v);
break;
}
if (s + s <= n) {
if (st.find(pr(s + s, s)) == st.end()) {
que.push(pr(pr(s + s, s), v + 1));
st.insert(pr(s + s, s));
}
}
if (s + m <= n) {
if (st.find(pr(s + m, m)) == st.end()) {
que.push(pr(pr(s + m, m), v + 1));
st.insert(pr(s + m, m));
}
}
}
return 0;
}

第三题

大模拟，细节考虑好就行。
由于构成数字的字符串只有四种，我比较讨巧的写了一个数字替代字符串的方法。#include <bits/stdc++.h>
using namespace std;
string str;
const string ans[4] = {"66666", "6...6", "....6", "6...."};
const int num[5][10] = {{0, 2, 0, 0, 1, 0, 0, 0, 0, 0},
{1, 2, 2, 2, 1, 3, 3, 2, 1, 1},
{1, 2, 0, 0, 0, 0, 0, 2, 0, 0},
{1, 2, 3, 2, 2, 2, 1, 2, 1, 2},
{0, 2, 0, 0, 2, 0, 0, 2, 0, 0}};
long long cheng(int &pos, int len) {
//printf("cheng(%d, %d)\n", pos, len);
long long res = 0;
while (pos < len) {
if (str[pos] == '6') {
res = res * 10 + 6;
pos++;
}
else if (str[pos] == '+') return res;

4000
else if (str[pos] == '-') return res;
else if (str[pos] == '*') {
pos++;
res *= cheng(pos, len);
}
}
return res;
}
long long jian(int &pos, int len) {
//printf("jian(%d, %d)\n", pos, len);
long long res = 0;
while (pos < len) {
if (str[pos] == '6') {
res = res * 10 + 6;
pos++;
}
else if (str[pos] == '+') {
return res;
}
else if (str[pos] == '-') {
return res;
}
else if (str[pos] == '*') {
pos++;
res *= cheng(pos, len);
}
}
return res;
}
long long cal(int &pos, int len) {
//printf("cal(%d, %d)\n", pos, len);
long long res = 0;
while (pos < len) {
if (str[pos] == '6') {
res = res * 10 + 6;
pos++;
}
else if (str[pos] == '+') {
pos++;
res += cal(pos, len);
}
else if (str[pos] == '-') {
pos++;
res -= jian(pos, len);
}
else if (str[pos] == '*') {
pos++;
res *= cheng(pos, len);
}
}
return res;
}
vector<int> v;
int main() {
int T, pos;
cin >> T;
while (T--) {
pos = 0;
cin >> str;
long long result = cal(pos, str.length());
v.clear();
if (result == 0) v.push_back(0);
else while (result) {
v.push_back(result % 10);
result /= 10;
}
for (int i = 0; i < v.size() / 2; i++) swap(v[i], v[v.size() - 1 - i]);
for (int j = 0; j < 5; j++) {
for (int i = 0; i < v.size(); i++) {
if (i > 0) cout << "..";
cout << ans[num[j][v[i]]];
}
cout << endl;
}
}
return 0;
}

第四题

很明显，先交换较小的数可以让平均数的变化较小。所以贪心即可#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e5 + 5;
long long a[MAX_N], b[MAX_N];
set<long long> sta, stb;
int main() {
int n, m, ans = 0;
long long suma = 0, sumb = 0, numa, numb;
scanf("%d%d", &n, &m);
for (int i = 0; i < n; i++) {
scanf("%lld", &a[i]);
sta.insert(a[i]);
suma += a[i];
}
for (int i = 0; i < m; i++) {
scanf("%lld", &b[i]);
stb.insert(b[i]);
sumb += b[i];
}
numa = n;
numb = m;
sort(a, a + n);
sort(b, b + m);
for (int ai = 0, bi = 0; ai < n || bi < m; ) {
if (ai == n) {
if (sta.find(b[bi]) == sta.end() && numb > 0) {
if ((suma + b[bi]) * numa > suma * (numa + 1)) {
if ((sumb - b[bi]) * numb > sumb * (numb - 1)) {
numa++;
suma += b[bi];
numb--;
sumb -= b[bi];
ans++;
}
}
}
bi++;
}
else if (bi == m) {
if (stb.find(a[ai]) == stb.end() && numa > 0) {
if ((sumb + a[ai]) * numb > sumb * (numb + 1)) {
if ((suma - a[ai]) * numa > suma * (numa - 1)) {
numa--;
suma -= a[ai];
numb++;
sumb += a[ai];
ans++;
}
}
}
ai++;
}
else if (a[ai] > b[bi]) {
if (sta.find(b[bi]) == sta.end() && numb > 0) {
if ((suma + b[bi]) * numa > suma * (numa + 1)) {
if ((sumb - b[bi]) * numb > sumb * (numb - 1)) {
numa++;
suma += b[bi];
numb--;
sumb -= b[bi];
ans++;
}
}
}
bi++;
}
else if (a[ai] < b[bi]) {
if (stb.find(a[ai]) == stb.end() && numa > 0) {
if ((sumb + a[ai]) * numb > sumb * (numb + 1)) {
if ((suma - a[ai]) * numa > suma * (numa - 1)) {
numa--;
suma -= a[ai];
numb++;
sumb += a[ai];
ans++;
}
}
}
ai++;
}
else {
ai++;
bi++;
}
}
printf("%d\n", ans);
return 0;
}

第五题

鉴于每次可以跳的范围是2H，所以我们每次弹跳二分的找到所有可跳的位置就行，然后就可以广搜了。#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e5;
bool vis[MAX_N * 2];
int n, T[MAX_N];
queue<int> que;
int erfen_big(int x) {
int L = 0, R = n - 1, ans = n - 1;
while (L <= R) {
int mid = (L + R) / 2;
if (T[mid] >= x) {
ans = mid;
R = mid - 1;
}
else {
L = mid + 1;
}
}
return ans;
}
int erfen_small(int x) {
int L = 0, R = n - 1, ans = 0;
while (L <= R) {
int mid = (L + R) / 2;
if (T[mid] <= x) {
ans = mid;
L = mid + 1;
}
else {
R = mid - 1;
}
}
return ans;
}
int main() {
int k, h, ans = 0;
scanf("%d%d%d", &n, &k, &h);
for (int i = 0; i < n; i++) {
scanf("%d", &T[i]);
}
sort(T, T + n);
que.push(0);
vis[0] = true;
while (!que.empty()) {
int x = que.front(); que.pop();
ans = max(ans, x);
int L = erfen_big(x - h), R = erfen_small(x + h);
//printf("x = %d, L = %d, R = %d\n", x, L, R);
for (int i = L; i <= R; i++) {
int y = 2 * T[i] - x;
if (y <= 0) continue;
if (vis[y]) continue;
que.push(y);
vis[y] = true;
}
}
printf("%d\n", ans);
return 0;
}