深信服C/C++笔试A卷
1.给一个数组和一个数字k。前两个比大小，小的放最后面，大的算赢一次，赢k次，输出该大的数。（ac）
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
vector<int> arr;
int tmp = 0;
while (cin >> tmp) {
arr.push_back(tmp);
}
int k = arr.back(), len = arr.size() - 1;
arr.resize(len);
if (len <= k) {
cout << *max_element(arr.begin(), arr.end());
return 0;
}
int count = 0;
int maxval = arr[0];
for (int i = 1; count < k; i++) {
if (maxval > arr[i]) {
count++;
arr.push_back(arr[i]);
}
else {
count = 1;
arr.push_back(maxval);
maxval = arr[i];
}
}
cout << maxval;
return 0;
} 2.最大排序连续序列（ac90%）
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
bool isright(vector<int> nums, int num, int i) {
sort(nums.begin(), nums.end() - i);
int j = 1;
for (; j < num - i; j++) {
if (nums[j] != nums[j - 1] + 1) break;
}
return j == num - i;
}
int main() {
int num = 0; cin >> num;
vector<int> nums(num);
for (int i = 0; i < num; i++) cin >> nums[i];
/*int num = 5;
vector<int> nums = { 3,1,2,4,6 };*/
for (int i = 0; i < num; i++) {
if (isright(nums, num, i)) {
cout << num - i;
return 0;
}
}
cout << 1;
return 0;
} 3.病毒侵入（ac，转化为区间即可）
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
int n = 0, x = 0; cin >> n >> x;
vector<int> nums(n);
for (int i = 0; i < n; i++) cin >> nums[i];
/*int n = 5,x = 4;
vector<int> nums = { 4,9,6,7,8 };*/
vector<vector<int>> arr;
for (int i = 0; i < n; i++) arr.push_back({ nums[i] - x, nums[i] + x });
vector<int> jiaoji = arr[0];
int count = 0;
for (int i = 1; i < n; i++) {
if (jiaoji[0] > arr[i][1] || jiaoji[1] < arr[i][0]) {
count++;
jiaoji = arr[i];
}
else {//有交集
jiaoji = { max(arr[i][0], jiaoji[0]), min(arr[i][1], jiaoji[1]) };
}
}
cout << count;
return 0;
}
#深信服笔试题##深信服##深信服校招#
