A Google Recruitment
热身题，素数，stoll函数，注意数据范围。
#include<iostream>
#include<string>
using namespace std;
bool check(const string &s) {
long long tmp = stoll(s);
if (tmp < 2)
return false;
for (int i = 2; (long long) i * i <= tmp; i++)
if (tmp % i == 0)
return false;
return true;
}
int main() {
int l, k;
string s;
cin >> l >> k;
cin >> s;
//这里要取等号
for (int i = 0; i + k <= s.length(); i++) {
string tmp = s.substr(i, k);
if (check(tmp)) {
cout << tmp;
return 0;
}
}
cout << "404";
return 0;
}
B Decode Registration Card of PAT
复杂的字符串题，需要细心，频繁输入输出的情况下使用scanf printf，这题使用cin cout最后一个样例TLE。
//高"IO"使用scanf
#include<iostream>
#include<vector>
#include<string>
#include<unordered_map>
#include<algorithm>
using namespace std;
const int MAXN = 10010;
struct node {
string id;
int score;
node() {
};
node(string a, int b) : id(a), score(b) {
};
//排序
bool operator<(const node &a) {
if (score != a.score)
return score > a.score;
else
return id < a.id;
}
} arr[MAXN];
int main() {
int n, m;
int tp;
string s;
cin >> n >> m;
for (int i = 0; i < n; i++)
cin >> arr[i].id >> arr[i].score;
for (int i = 1; i <= m; i++) {
cin >> tp >> s;
printf("Case %d: %d %s\n", i, tp, s.c_str());
if (tp == 1) {
vector<node> ans;
for (int i = 0; i < n; i++)
//匹配考试等级
if (arr[i].id[0] == s[0])
ans.push_back(arr[i]);
if (!ans.size()) {
printf("NA\n");
continue;
}
sort(ans.begin(), ans.end());
for (const node &a:ans)
printf("%s %d\n", a.id.c_str(), a.score);
} else if (tp == 2) {
int cnt = 0, res = 0;
for (int i = 0; i < n; i++)
//匹配考场号
if (arr[i].id.substr(1, 3) == s) {
++cnt;
res += arr[i].score;
}
if (!cnt) {
printf("NA\n");
continue;
}
printf("%d %d\n", cnt, res);
} else {
unordered_map<string, int> map;
for (int i = 0; i < n; i++) {
string tmp = arr[i].id.substr(1, 3);
//匹配日期，但是要对考场号处理
if (arr[i].id.substr(4, 6) == s) {
if (map.find(tmp) == map.end()) {
map[tmp] = 0;
}
map[tmp]++;
}
}
vector<node> ans;
for (auto i:map) {
ans.push_back(node(i.first, i.second));
}
if (!ans.size()) {
printf("NA\n");
continue;
}
sort(ans.begin(), ans.end());
for (const node &a:ans)
printf("%s %d\n", a.id.c_str(), a.score);
}
}
return 0;
}
C Vertex Coloring
简单图论，邻接表存边，然后按照题意编程即可。
//邻接表存储，否则TLE
#include<iostream>
#include<vector>
#include<unordered_set>
#include<algorithm>
#define ac cin.tie(0);cin.sync_with_stdio(0);
using namespace std;
const int MAXN = 10010;
vector<int> G[MAXN];
int arr[MAXN];
int n, m, k;
int main() {
ac
int a, b;
cin >> n >> m;
while (m--) {
cin >> a >> b;
//无向图，但是只需要存储"半条边"，因为遍历的时候从0-n
if (a > b)
swap(a, b);
G[a].push_back(b);
}
cin >> k;
while (k--) {
bool op = true;
//记录颜色数
unordered_set<int> set;
for (int i = 0; i < n; i++)
cin >> arr[i];
for (int i = 0; i < n; i++) {
set.insert(arr[i]);
for (int j:G[i])
//邻接又颜色相等，不符合题意
if (arr[i] == arr[j]) {
op = false;
break;
}
}
if (op)
cout << set.size() << "-coloring";
else
cout << "No";
if (k)
cout << endl;
}
return 0;
}
D Heap Paths
中等难度的树题，如果不是完全二叉树就有难度了，DFS。
#include<cstdio>
#include<vector>
using namespace std;
const int MAXN = 1010;
int arr[MAXN];
vector<int> path;
int n;
bool op1 = true, op2 = true;
void dfs(int root, int fa) {
if (fa != -1) {
//子节点比父结点大，不可能是大根堆
if (arr[root] > arr[fa])
op1 = false;
//子节点比父结点小，不可能是小根堆
if (arr[root] < arr[fa])
op2 = false;
}
//加入path
path.push_back(arr[root]);
if (2 * root > n) {
for (int i = 0; i < path.size(); i++) {
if (i)
printf(" ");
printf("%d", path[i]);
}
printf("\n");
//回溯 pop
path.pop_back();
return;
}
//先遍历右子树，完全二叉树右儿子为2*root+1
if (2 * root + 1 <= n)
dfs(2 * root + 1, root);
//再遍历左子树，完全二叉树左儿子为2*root
dfs(2 * root, root);
//回溯pop
path.pop_back();
}
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &arr[i]);
dfs(1, -1);
if (op1)
printf("Max Heap");
else if (op2)
printf("Min Heap");
else
printf("Not Heap");
return 0;
}
PS.明天PAT冬季赛，冲冲冲！！！