D-小红的因式分解_牛客周赛 Round 50 (nowcoder.com)
思路:
巨蠢的题目,ax^2+bx+c=a1*a2*x^2+(b1*a2+b2*a1)x+b1*b2,即:
a=a1*a2,b=a1*b2+a2*b1,c=b1*b2
数据范围很小,直接暴力枚举吧(注意条件)
代码:
cpp
void solve()
{
ll a, b, c, a1, a2, b1, b2;
cin >> a >> b >> c;
for (int a1 = -1000; a1 <= 1000; a1++)
{
if (a1 == 0 || a % a1 != 0)
continue;
for (int b1 = -1000; b1 <= 1000; b1++)
{
if (b1 == 0 || c % b1 != 0)
continue;
a2 = a / a1, b2 = c / b1;
if (a1 * b2 + a2 * b1 == b)
{
cout << a1 << " " << b1 << " " << a2 << " " << b2 << endl;
return;
}
}
}
cout << "NO" << endl;
return;
}
思路:
仔细看题目,题目的意思是有可能,即没可能的情况只有一种(即x1,y1和x2,y2相交的情况)
AC代码:
cpp
#define _CRT_SECURE_NO_WARNINGS 1
//------ 棘手大学 世界第一 ------//
#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<cmath>
using namespace std;
//gcd最大公约数,lcm最小公倍数
typedef long long ll;
#define IOS ios::sync_with_stdio(false); cin.tie(nullptr), cout.tie(nullptr);
priority_queue<int, vector<int>, greater<int>> pq;
map<int, int>mp;
const int N = 2e6 + 10;
int mod(string a, ll b)//高精度a除以单精度b
{
ll n = 0;
for (int i = 0; i < a.size(); i++) n = (n * 10 + (a[i] - '0')) % b; //求出余数
return n;
}
string gcd(string a, ll b) // 欧几里得算法
{
while (b != 0)
{
ll temp = b;
b = mod(a, b);
a = to_string(temp);
}
return a;
}
inline int gcd1(int a, int b)//a、b不可以为0(很快)
{
while (b ^= a ^= b ^= a %= b);
return a;
}
inline int gcd2(int a, int b)//a、b可以为0(很快)
{
if (b) while ((a %= b) && (b %= a));
return a + b;
}
int lcm1(int a, int b) {
return a * b / gcd1(a, b);
}
void solve()
{
ll x1, y1, x2, y2;
cin >> x1 >> y1;
cin >> x2 >> y2;
if (x1 < y1)
{
if (x2 >= y2)
{
cout << "NO" << endl;
return;
}
else
{
cout << "YES" << endl;
return;
}
}
else if (x1 > y1)
{
if (x2 <= y2)
{
cout << "NO" << endl;
return;
}
else
{
cout << "YES" << endl;
return;
}
}
else
{
cout << "YES" << endl;
return;
}
}
int main()
{
IOS;
int t;
cin >> t;
while(t--)
solve();
return 0;
}
思路:
不会,只能看题解写了(QWQ)
AC代码:
cpp
#include<bits/stdc++.h>
using namespace std;
void solve() {
int x, y, k;
cin >> x >> y >> k;
while (k != 0 && x != 1) {
int t = min(k, y - x % y);
x += t;
k -= t;
while (x % y == 0) {
x /= y;
}
}
k %= (y - 1);
x = x + k;
cout << x << endl;
}
int main() {
int t;
cin >> t;
while (t--)
solve();
return 0;
}
abc363只a了三道题
几个if语句就能过
cpp
void solve()
{
ll n;
cin >> n;
if (n <= 1 && n <= 99)
{
cout << 100 - n << endl;
return;
}
if (n >= 100 && n <= 199)
{
cout << 200 - n << endl;
return;
}
if (n >= 200 && n <= 299)
{
cout << 300 - n << endl;
return;
}
if (n >= 300 && n <= 399)
{
cout << 399 - n << endl;
return;
}
}
B - Japanese Cursed Doll (atcoder.jp)
思路:
判断当前有多少个>=p的就行,不满足要求的时候我们从最大的开始看(其实就是求有限范围里满足题目要求的最小值)
AC代码:
cpp
ll l[110];
bool cmp(ll a, ll b)
{
return a > b;
}
void solve()
{
ll n, t, p;
cin >> n >> t >> p;
ll k = 0;
for (int i = 1; i <= n; i++)
{
cin >> l[i];
if (l[i] >= t)
k++;
}
sort(l + 1, l + 1 + n, cmp);
ll ans = 0, tt = k;
if (k >= p)
{
cout << "0" << endl;
return;
}
for (int i = 1; i <= n; i++)
{
if (tt == p)
{
break;
}
if (l[i] < t)
{
ans = max(ans, t - l[i]);
tt++;
}
}
cout << ans << endl;
return;
}