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9.4 网易后端笔试-题解

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2023-03-28

9.4 网易后端笔试-题解

第一题 求两个数的最小差

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n;
int a[200005];

int main() {
cin >> n;

for (int i = 1; i <= n; i++) {
cin >> a[i];
}

sort(a + 1, a + n + 1);

int minSub = 0x3f3f3f3f;
int ans1, ans2;
for (int i = 2; i <= n; i++) {
minSub = min(a[i] - a[i - 1], minSub);
}

cout << minSub << endl;

return 0;
}

第二题 求乘积最大的数字拆分方式,枚举等分的份数即可

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n;

int gcdMath(int x, int y) {
return x % y == 0 ? y : gcdMath(y, x % y);
}

int main() {
cin >> n;

if (n <= 4) {
cout << n << ' ' << 1 << endl;
return 0;
}

int i, last = 0;
for (i = 2; i < n; i++) {
double tmp = (double)n / i;
double now = 1.0;
for (int j = 1; j <= i; j++) {
now *= tmp;
}
if (now < last) {
i--;
break;
}
else {
last = now;
}
}

int up = 1, down = 1;
for (int k = 1; k <= i; k++) {
up *= n;
}
for (int k = 1; k <= i; k++) {
down *= i;
}

int gcd = gcdMath(up, down);
up /= gcd;
down /= gcd;

cout << up << ' ' << down << endl;

return 0;
}

第三题 求两个数组的差分数列数组的最长公共子数组,遍历一遍即可

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

int n;
int a[200005];
int b[200005];

int main() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
for (int i = 1; i <= n; i++) {
cin >> b[i];
}

int ans = 0, idx = 1;
while (idx < n) {
int count = 0;
while (idx < n && a[idx + 1] - a[idx] == b[idx + 1] - b[idx]) {
count++;
idx++;
}
ans = max(ans, count + 1);
idx++;
}

cout << ans << endl;

return 0;
}

第四题 可以确定的涂色块,二分搞一搞,看看方格一定会落到哪个颜色中

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <vector>
using namespace std;

int a[1000005], sum[1000005];

int main() {
int w, n, cur = 0;
cin >> w >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
sum[i] = sum[i - 1] + a[i];
if (i != 1) {
sum[i]++;
}
}
vector<int> vc;
for (int i = 1; i <= w; i++) {
int pos = lower_bound(sum, sum + n + 1, i) - sum;
pos--;
int res = sum[n] - sum[pos];
if (pos != 0) {
res--;
}
if (w - i < res) {
vc.push_back(i);
}
}
cout << vc.size() << endl;
for (int i = 0; i < vc.size(); i++) {
if (i == vc.size() - 1) {
cout << vc[i] << endl;
}
else {
cout << vc[i] << " ";
}
}
return 0;
}
#网易##笔试#
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