文章目录
11、构造
11.1、小浩的ABC
c
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
using ll = long long;
int main(){
IOS;
int t;cin>>t;
while(t--){
ll x;cin>>x;
// A B C <=10^6
if(x<2){
cout<<"-1"<<'\n';continue;
}
if(x<=1000000){
cout<<x-1<<" 1 1"<<'\n';continue;
}
ll a=1000000,c=x%a,b;
if(c==0)c=a;
b=(x-c)/a;
cout<<a<<' '<<b<<' '<<c<<'\n';
}
return 0;
}11.2、小新的质数序列挑战
c
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
using ll = long long;
int main(){
IOS;
int T;cin>>T;
while(T--){
//质数序列，只能被1或本身整除的序列
ll a,b;cin>>a>>b;
if(a==1||b==1){
cout<<"-1\n";continue;
}
auto gcd = [&](ll a,ll b){
while(b!=0){
ll t=a%b;
a=b;
b=t;
}
return a;
};
ll g=gcd(a,b);
if(g==1){
cout<<"1\n";
}else{
cout<<"0\n";
}
}
return 0;
}11.3、小蓝找答案
c
#include <bits/stdc++.h>
using LL = long long;
using Pair = std::pair<int, int>;
#define inf 1'000'000'000
void solve(const int &Case) {
int n;
std::cin >> n;
std::vector<int> a(n);
for (auto &x: a)std::cin >> x;
auto ck = [&]() {
for (int i = 1; i < n; i++) {
if (a[i] <= a[i - 1])return false;
}
return true;
};
if (ck()) {
std::cout << "1\n";
return;
}
int l = 2, r = n, ret = n;
while (l <= r) {
int mid = (l + r) >> 1;
std::vector<Pair> A;
int flag = 0;
std::function<void(int)> push = [&](int x) {
if (x <= 0) {
flag = 1;
return;
}
while (!A.empty() && A.back().first > x)A.pop_back();
if (A.empty()) {
A.emplace_back(x, 1);
return;
}
if (A.back().first == x) {
A.back().second++;
if (A.back().second == mid) {
push(x - 1);
A.emplace_back(x, 0);
}
}
else {
A.emplace_back(x, 1);
}
};
A.emplace_back(a[0], 0);
for (int i = 1; i < n; i++) {
if (a[i] > a[i - 1])A.emplace_back(a[i], 0);
else push(a[i]);
if (flag)break;
}
if (flag)l = mid + 1;
else {
ret = mid;
r = mid - 1;
}
}
std::cout << ret << '\n';
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
int T = 1;
// std::cin >> T;
for (int Case = 1; Case <= T; Case++)solve(Case);
return 0;
}11.4、小蓝的无限集
c
#include <bits/stdc++.h>
using LL = long long;
using Pair = std::pair<int, int>;
#define inf 1'000'000'000
void solve(const int &Case) {
int a, b, n;
std::cin >> a >> b >> n;
if (a == 1) {
if (n % b == 1)std::cout << "Yes\n";
else std::cout << "No\n";
return;
}
// 枚举 a ^ i
LL pw = 1;
while (pw <= n) {
if ((n - pw) % b == 0) {
std::cout << "Yes\n";
return;
}
pw *= a;
}
std::cout << "No\n";
}
int main() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
int T = 1;
std::cin >> T;
for (int Case = 1; Case <= T; Case++)solve(Case);
return 0;
}12、高精度
12.1、阶乘数码(高精度*单精度)
c
#include<bits/stdc++.h>
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
void mult(vector<int>& a,int b){
int d=0;
for(int i=0;i<a.size();i++){
int p=a[i]*b+d;
a[i]=p%10;
d=p/10;
}
while(d){
a.push_back(d%10);
d/=10;
}
}
int main() {
IOS;
int t;cin>>t;
while(t--){
int n,a1;cin>>n>>a1;
vector<int> ans(1,1);
for(int i=2;i<=n;i++){
mult(ans,i);//n的阶乘
}
int res=0;
for(const auto &x:ans){
if(x==a1)res++;
}
cout<<res<<"\n";
}
return 0;
}越来越难了。。。。。