PAT 2022春季乙级 C++ 满分题解
PAT 2022春季乙级 C++ 满分题解
7-1暴力破解（15分）
算法标签 排序 枚举
AC代码
#include
using namespace std;
const int N = 10;
int a[N];
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n;
cin>>n;
for(int i=0;i>a[i];
}
sort(a,a+n);
int cnt=0;
for(int i=0;i 7-2 学霸 (20 分)
算法标签 结构体 排序 模拟
AC代码
#include
#include
#include
using namespace std;
const int MAXN = 1e6;
struct node {
int num; //编号
int period; //学时
int cnt; //门数
}per[MAXN];
bool cmp(struct node a, struct node b){
if(a.period != b.period)
return a.period > b.period;
else if (a.period == b.period && a.cnt != b.cnt)
return a.cnt > b.cnt;
else if (a.period == b.period && a.cnt == b.cnt)
return a.num < b.num;
}
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int N, peopleNum, period;
cin >> N;
for (int i = 0; i < N; i++){
cin >> period >>peopleNum;
for (int j = 0; j < peopleNum; j++) {
int temp;
cin >> temp;
per[temp].num = temp;
per[temp].cnt++;
per[temp].period += period;
}
}
sort(per, per + MAXN, cmp);
cout << per[0].period << " " << per[0].cnt << endl;
cout << setfill('0') << setw(5) << per[0].num;
for (int i = 1; i < MAXN; i++) {
if (per[i].cnt == per[0].cnt && per[i].period == per[0].period) {
cout << " " << per[i].num;
}
}
return 0;
}
7-3 排课 (20 分)
算法标签 结构体 模拟
AC代码
#include
#define int long long
using namespace std;
const int MAXN = 1e4 + 7;
struct node {
int L;
set time_t;
int time;
int room;
int cnt = 0;//本班级排课与老师不冲突
}per[MAXN];
signed main(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int all = 0;
int N, M, K;
cin >> N >> M >> K;
for (int i = 0; i < N; i++) {
cin >> per[i].L;
for (int j = 0; j < per[i].L; j++) {
int temp;
cin >> temp;
per[i].time_t.insert(temp);
}
cin >> per[i].time >> per[i].room;
//处理两个班之间的 跟所有的班级内排课不冲突的班比较
if (i)
{
for (int m = 0; m < i; m++) {
if (per[i].time == per[m].time && per[i].room == per[m].room && !per[m].cnt) {
cout << "ERROR: Conflict between " << i + 1 << " and " << m + 1 << "." << endl;
per[i].cnt = 1;
all++;
}
}
}
//处理班内的
if (per[i].time_t.find(per[i].time) != per[i].time_t.end())
{
cout << "ERROR: Conflict with instructor for " << i + 1 << "." << endl;
per[i].cnt = 1;
all++;
}
}
if (!all) cout << "Perfect Arrangement for " << N << " classes!";
return 0;
}
考场上认为比较复杂 没有模拟出来，下来仔细想想，本质为循环嵌套查找
7-4 简易测谎 (20 分)
算法标签 字符串 模拟 排序
AC代码
#include
#define int long long
using namespace std;
const int N = 105;
string check(string s, int t){
int cnt=0;
// 条件1
if(s[0]=='f'){
cnt-=2;
}
// 条件2
if(s[s.size()-1]=='a'){
cnt-=1;
}
vector l(N, 0);
vector ll(N, 0 );
int len=1;
// 条件3
for(int i=0;i());
sort(ll.begin(),ll.end(), greater());
for(int i=0;i5){
cnt+=3;
}
else{
break;
}
}
for(int i=0;i3){
cnt+=5;
}
else{
break;
}
}
if(cnt>t){
return to_string(cnt)+"!!!";
}
else{
return to_string(cnt);
}
}
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n,t,k;
cin>>n>>t>>k;
while(k--){
string s;
cin>>s;
cout< 考场上认为比较复杂 没有模拟出来 下来理清思路，依次模拟
7-5 前K大数 (25 分)
算法标签 队列
思路
由于本题空间限制严格，直接排序导致内存超限，可以采用优先队列优化，优先队列只存储前5大的值即可。
AC代码
#include
#define int long long
using namespace std;
const int INF = -0x3f3f3f3f, K = 5;
int a[K] = {INF};
priority_queue, greater> q;
signed main(){
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n,k;
// cin 容易STL
// cin>>n>>k;
scanf("%lld%lld", &n,&k);
if(nK){
q.pop();
}
}
int p=0;
while (!q.empty()){
a[p++] = q.top();
q.pop();
}
for (int i = p - 1, j = 0; j < k; i--, j++) {
if (i == p - 1) {
printf("%d", a[i]);
}
else {
printf(" %d", a[i]);
}
}
return 0;
}
考场上用sort，只得了13分
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