美团笔试统计&题解 2022-08-06 后端&数开&软开
1.包装礼盒
2.划分数组
3.反转朝上数字个数大于一半
4.训练集测试集划分
5.拼接字符串中的第k个字符
#include <bits/stdc++.h>
using namespace std;
int fun(int x, int y) {
if(x == 0 || y == 0) return 0;
if(x == 1 && y == 1) return 0;
if(x > y) swap(x, y);
int z = y / 2;
if(z >= x) {
return x;
} else {
return (x + y) / 3;
}
}
int main() {
int T;
cin >> T;
while(T--) {
int x, y;
cin >> x >> y;
cout << fun(x, y) << endl;
}
return 0;
} 2.前缀和、后缀和 #include <bits/stdc++.h>
using namespace std;
int a[100005];
int b[100005];
int c[100005];
int main() {
int n;
cin >> n;
b[0] = c[n] = 0;
for(int i = 0; i < n; i++) {
cin >> a[i];
if(a[i] >= 0) {
b[i + 1] = b[i] + 1;
} else {
b[i + 1] = b[i];
}
}
for(int i = n - 1; i >= 0; i--) {
if(a[i] <= 0) {
c[i] = c[i + 1] + 1;
} else {
c[i] = c[i + 1];
}
}
int ans = c[0];
for(int i = 0; i < n; i++) {
ans = min(ans, b[i + 1] + c[i + 1]);
}
cout << ans << endl;
return 0;
} 3.记录反面每个数出现的个数，即反转后是该数的个数，然后遍历正面每个数的情况并统计，特殊情况是，正面的数都不满足，就得再看看反面的 #include <bits/stdc++.h>
using namespace std;
int a[100005];
int b[100005];
int main() {
int n;
cin >> n;
unordered_map<int, int>mp;//记录反转后是x的数个数
for(int i = 0; i < n; i++) {
cin >> a[i];
}
for(int i = 0; i < n; i++) {
cin >> b[i];
if(b[i] != a[i]) {
mp[b[i]]++;
}
}
int target;
if(n & 1) target = n / 2 + 1;
else target = n / 2;
sort(a, a + n);
int ans = INT_MAX;
for(int i = 0; i < n; i++) {
int count = 0;
int j = i;
while(j < n && a[i] == a[j]) {
count++;
j++;
}
if(count >= target) {
ans = 0;
break;
}
if(mp[a[i]] >= target - count) {
ans = min(ans, target - count);
}
i = j - 1;
}
if(ans != INT_MAX) cout << ans << endl;
else {
for(int i = 0; i < n; i++) {//正面的数字都不能满足要求，就只能看看能不能反转target个反面的数字
if(mp[b[i]] >= target) {
ans = target;
break;
}
}
if(ans != INT_MAX) cout << ans << endl;
else cout << -1 << endl;
}
return 0;
} 4.排序 #include <bits/stdc++.h>
using namespace std;
int a[10005];
int main() {
int n, k;
cin >> n >> k;
vector<vector<int>>v(k);
for(int i = 0; i < n; i++) {
cin >> a[i];
v[a[i] - 1].push_back(i + 1);
}
vector<int>train;
vector<int>test;
for(int i = 0; i < k; i++) {
int m = ceil(1.0 * v[i].size() / 2);
for(int j = 0; j < m; j++) {
train.push_back(v[i][j]);
}
for(int j = m; j < v[i].size(); j++) {
test.push_back(v[i][j]);
}
}
sort(train.begin(), train.end());
sort(test.begin(), test.end());
for(int i = 0; i < train.size(); i++) {
if(i == train.size() - 1) cout << train[i] << endl;
else cout << train[i] << " ";
}
for(int i = 0; i < test.size(); i++) {
if(i == test.size() - 1) cout << test[i] << endl;
else cout << test[i] << " ";
}
return 0;
} 5.二分+dfs，预处理所有的字符串长度，分前半部分、后半部分和wow分别处理 #include <bits/stdc++.h>
using namespace std;
vector<long long>count_;
string s = "MeiTuan";
char dfs(long long k, long long cur_len) {
// cout << k << endl;
if(k <= s.size()) return s[k - 1];
// cout << count_[i] << endl;
if(k == cur_len || k == cur_len - 2) return 'w';
if(k == cur_len - 1) return 'o';
long long next_len = (cur_len - 3) / 2;
if(k > next_len) {//k在后半部分
k -= next_len;
return dfs(next_len - k + 1, next_len);
}
return dfs(k, next_len);
}
int main() {
long long x = 7;
long long y = LLONG_MAX;
while(true) {
count_.push_back(x);
if(x >= (y - 3) / 2) break;
x = x * 2 + 3;
}
int T;
cin >> T;
while(T--) {
long long k;
cin >> k;
int i = upper_bound(count_.begin(), count_.end(), k) - count_.begin();
long long cur_len = count_[i];
cout << dfs(k, cur_len) << endl;
}
return 0;
} 
