字节5.6笔试 客户端题解
第一题
第三题
第四题， #实习##笔试题目##C/C++##字节跳动#
buff总共持续时间，新buff会重置旧buff
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int n;
cin >> n;
vector<pair<int, int>> pa(n);
for (int i = 0; i < n; ++i) {
cin >> pa[i].first;
}
for (int i = 0, tmp; i < n; ++i) {
cin >> tmp;
pa[i].second = pa[i].first + tmp - 1;
}
sort(pa.begin(), pa.end());
int res = 0;
int start = pa[0].first, last = pa[0].second;
for (int i = 1; i < n; ++i) {
if (last < pa[i].first) {
res += last - start + 1;
}
else {
res += pa[i].first - start;
}
start = pa[i].first;
last = pa[i].second;
}
res += last - start + 1;
cout << res;
} 第二题
有效数字集，匹配数字前缀成功即有效
主要会有溢出问题，用long long #include <iostream>
#include <vector>
#include <unordered_set>
using namespace std;
using ll = long long ;
int main () {
int N;
cin >> N;
for (int i = 0; i < N; ++i) {
unordered_set<ll> mset;
int T;
cin >> T;
vector<ll> data(T);
for (int j = 0; j < T; ++j) {
cin >> data[j];
mset.insert(data[j]);
}
bool flag = false;
for (int j = 0; j < T; ++j) {
ll val = data[j] / 10;
while (val) {
if (mset.count(val)) {
flag = true;
break;
}
else {
val /= 10;
}
}
if (flag) break;
}
if (flag) cout << "YES" << endl;
else cout << "NO" << endl;
}
} 完成两个任务，总共可支配时间m，给出n个任务，求能完成的最大价值
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int n, m;
cin >> n >> m;
vector<pair<int, int>> pa(n);
for (int i = 0; i < n; ++i) {
cin >> pa[i].second >> pa[i].first; //时间 价值
}
sort(pa.rbegin(), pa.rend());//按照价值降序
int res = 0;
for (int i = 0; i < n; ++i) {
if (pa[i].second >= m) continue;
int tmpSum = 0;
for (int j = i + 1; j < n; ++j) {
tmpSum = pa[i].first + pa[j].first;
if (tmpSum <= res) break;
if (pa[i].second + pa[j].second <= m) res = max(res, tmpSum);
}
}
cout << res;
} 第四题，
删除一个连续子数组，求新数组的最大严格上升子数组
只过了35，求大佬补充 
