2025寒假天梯赛训练1
2025寒假天梯赛训练1
7-1 心理阴影面积
思路
用一半的面积减去一个梯形和一个三角形面积即可。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int a, b;
cin >> a >> b;
cout << (5000 - (100 - a) * 50 - b * 50) << "\n";
return 0;
}
7-2 人与神
思路
直接输出即可。
代码
To iterate is human, to recurse divine.
7-3 通讯录的录入与显示
思路
按题意输出即可。
注意记录编号可能为负数。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<array<string,5>> s(n);
for(auto &x : s){
for(int i = 0;i < 5;i ++){
cin >> x[i];
}
}
int k;
cin >> k;
while(k--){
int x;
cin >> x;
if(x >= n || x < 0){
cout << "Not Found\n";
}else{
cout << s[x][0] << " " << s[x][3] << " " << s[x][4] << " " << s[x][2] << " " << s[x][1] << "\n";
}
}
return 0;
}
7-4 算术入门之加减乘除
思路
按题意要求输出即可。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int a,b;
cin >> a >> b;
cout << a << " + " << b << " = " << a + b << "\n" ;
cout << a << " - " << b << " = " << a - b << "\n" ;
cout << a << " * " << b << " = " << a * b << "\n" ;
if(a % b == 0){
cout << a << " / " << b << " = " << a / b << "\n" ;
}else{
cout << fixed << setprecision(2) << a << " / " << b << " = " << a * 1.0 / b << "\n" ;
}
return 0;
}
7-5 出生年
思路
按题意模拟。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int a,b;
cin >> a >> b;
int cs = a;
while(1){
set<int> has;
if(a < 1000){
has.insert(0);
}
int t = a;
while(t){
has.insert(t%10);
t/=10;
}
if(has.size() == b){
printf("%d %04d\n",a-cs,a);
break;
}
a ++;
}
return 0;
}
7-6 九宫格输入法
思路
按题意模拟。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
vector<string> s{
"0 ",
"1,.?!",
"2ABC",
"3DEF",
"4GHI",
"5JKL",
"6MNO",
"7PQRS",
"8TUV",
"9WXYZ"
};
string x;
while (cin >> x) {
int a = x[0] - '0';
int b = x.size();
b--;
b %= (int)s[a].size();
cout << s[a][b];
}
return 0;
}
7-7 螺旋方阵
思路
按题意模拟。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector a(n + 1, vector(n + 1, 0));
int has = 0, x = 1, y = 0;
while (has < n * n) {
while(y + 1 <= n && !a[x][y + 1]){
y ++;
a[x][y] = ++has;
}
while(x + 1 <= n && !a[x + 1][y]){
x ++;
a[x][y] = ++has;
}
while(y - 1 >= 1 && !a[x][y - 1]){
y --;
a[x][y] = ++has;
}
while(x - 1 >= 1 && !a[x-1][y]){
x --;
a[x][y] = ++has;
}
}
for(int i = 1;i <= n;i ++){
for(int j = 1;j <= n;j ++){
cout << setw(3) << a[i][j];
}
cout << "\n";
}
return 0;
}
7-8 抓老鼠啊~亏了还是赚了？
思路
有点需要注意细节的模拟。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string s;
cin >> s;
int happy = 0, nohappy = 0, sad = 0;
int nai = 0, now = 0;
string ans = "";
auto clear = [&]() {
happy--;
happy = max(happy, 0);
nohappy--;
nohappy = max(nohappy, 0);
sad--;
sad = max(sad, 0);
};
for (int i = 0; i < s.size() - 1; i ++) {
if (s[i] == 'X') {
if (happy || (!nohappy && !sad)) {
ans += "U";
nohappy = 2;
} else {
ans += "-";
}
} else if (s[i] == 'T') {
if (happy || (!nohappy && !sad)) {
ans += "D";
now += 7;
sad = 3;
} else {
ans += "-";
}
} else {
if (happy || (!nohappy && !sad)) {
ans += "!";
now -= 3;
happy = 3;
} else {
ans += "-";
}
}
clear();
}
cout << ans << "\n" << now << "\n" ;
return 0;
}
7-9 Windows消息队列
思路
用优先队列维护优先级即可。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
priority_queue<pair<int, string>,vector<pair<int, string>>,greater<pair<int, string>>> Q;
while (n--) {
string x;
cin >> x;
if (x == "GET") {
if (Q.empty()) {
cout << "EMPTY QUEUE!\n";
} else {
auto [_, s] = Q.top();
Q.pop();
cout << s << "\n";
}
} else {
int m;
cin >> x >> m;
Q.emplace(m, x);
}
}
return 0;
}
7-10 名人堂与代金券
思路
考察排序。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n,g,k;
cin >> n >> g >> k;
vector<pair<string,int>> a(n);
int need = 0;
for(auto &[x,y] : a){
cin >> x >> y;
if(y >= g){
need += 50;
}else if(y >= 60){
need += 20;
}
}
cout << need << "\n";
sort(a.begin(),a.end(),[](auto x,auto y){
if(x.second != y.second) return x.second > y.second;
return x.first < y.first;
});
int rank = 1;
for(int i = 0;i < n;i ++){
if(rank > k) break;
cout << rank << " " << a[i].first << " " << a[i].second << "\n";
while(i + 1 < n && a[i + 1].second == a[i].second){
i ++;
cout << rank << " " << a[i].first << " " << a[i].second << "\n";
}
rank = i + 2;
}
return 0;
}
7-11 用扑克牌计算24点
思路
答辩题。
实际要注意的细节比较多，括号的顺序可以直接列出来然后挨个判断，其余的就是暴力搜索了。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string a[4];
for (int i = 0; i < 4; i ++) {
cin >> a[i];
}
auto check = [&](string & s)->bool{
int i = 0;
auto calc = [](double x, double y, char op)->double{
if (op == '+') return x + y;
if (op == '-') return x - y;
if (op == '*') return x * y;
return x * 1.0 / y;
};
auto Calc = [&](auto && self)->double{
char op = '+';
stack<double> st;
for (; i < s.size(); i++) {
if (s[i] == '(') {
i ++;
st.push(self(self));
} else if (s[i] != ')') {
if (isdigit(s[i])) {
int x = s[i] - '0';
if (isdigit(s[i + 1])) {
x = x * 10 + s[i + 1] - '0';
i ++;
}
st.push(x);
} else {
op = s[i];
}
} else {
double y = st.top();
st.pop();
double x = st.top();
st.pop();
return calc(x, y, op);
}
}
if (st.size() > 1) {
double y = st.top();
st.pop();
double x = st.top();
st.pop();
st.push(calc(x, y, op));
}
return st.top();
};
bool ok = Calc(Calc) == 24;
return ok;
};
string p = "+-*/";
string s = "((2-(3+12))+12)";
sort(a, a + 4);
do {
for (auto x : p) {
for (auto y : p) {
for (auto z : p) {
s = "((" + a[0] + x + a[1] + ")" + y + a[2] + ")" + z + a[3];
if (check(s)) {
cout << s << "\n";
return 0;
}
s = "(" + a[0] + x + a[1] + ")" + y + "(" + a[2] + z + a[3] + ")";
if (check(s)) {
cout << s << "\n";
return 0;
}
s = a[0] + x + "(" + a[1] + y + "(" + a[2] + z + a[3] + "))" ;
if (check(s)) {
cout << s << "\n";
return 0;
}
s = a[0] + x + "((" + a[1] + y + a[2] + ")" + z + a[3] + ")";
if (check(s)) {
cout << s << "\n";
return 0;
}
s = "(" + a[0] + x + "(" + a[1] + y + a[2] + "))" + z + a[3];
if (check(s)) {
cout << s << "\n";
return 0;
}
}
}
}
} while (next_permutation(a, a + 4));
cout << "-1\n";
return 0;
}
7-12 玩转二叉树
思路
先用中序前序按层还原二叉树，然后对于每一层反向输出即可。
代码
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<int> pre(n), mid(n);
for (int i = 0; i < n; i ++) {
cin >> mid[i];
}
for (int i = 0; i < n; i ++) {
cin >> pre[i];
}
map<int, vector<int>> tr;
int len = 0;
auto dfs = [&](auto && self, int rt, int st, int ed, int len)->void{
if (st > ed) {
return;
}
int root = st;
while (root < ed && pre[rt] != mid[root]) {
root ++;
}
self(self, rt + 1, st, root - 1, len + 1);
self(self, rt + root - st + 1, root + 1, ed, len + 1);
tr[len].emplace_back(pre[rt]);
};
dfs(dfs, 0, 0, n - 1, 0);
vector<int> ans;
for (int i = 0; i < n; i ++) {
if (tr[i].size()) {
for (int j = tr[i].size() - 1; j >= 0; j --) {
ans.push_back(tr[i][j]);
}
}
}
for (int i = 0; i < n; i ++) {
cout << ans[i] << " \n"[i == n - 1];
}
return 0;
}
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