A - Cats and Dogs
#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
int a, b, c;
int main(){
cin >> a >> b >> c;
if (c >= a && c <= a + b) cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
B - Toll Gates
长度为n+1的路上有m个收费站,每个收费站收费1元,现在从位置x出发,问走到0或者走到n位置,哪个方案收费最少
直接扫一遍即可
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
int n, m, x, a[N];
int main() {
cin >> n >> m >> x;
for (int i = 1; i <= m; i++) {
int x;
cin >> x;
a[x] = 1;
}
int tmp = 0;
for (int i = 0; i <= x; i++) {
tmp += a[i];
}
int res = tmp;
tmp = 0;
for (int i = n; i >= x; i--) {
tmp += a[i];
}
res = min(res, tmp);
cout << res << endl;
return 0;
}
C - Many Medians
给出n个数(n为偶数),要求对于每个数,都输出除了它以外,n-1个数的中位数
首先求出这n个数排序后最中间的两个数,答案肯定是这两个数中的一个
假设小的那个是res1,大的是res2,那么对于每个数来说,如果这个数小于等于res1,那么去除掉它后,中位数肯定是res2
反之,答案为res1
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
int n, a[N], b[N];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) cin >> a[i], b[i] = a[i];
sort(a + 1, a + 1 + n);
int res1 = a[n / 2], res2 = a[n / 2 + 1];
for (int i = 1; i <= n; i++) {
if (b[i] <= res1)
cout << res2 << endl;
else
cout << res1 << endl;
}
return 0;
}
D - Binomial Coefficients
给出n个数,要求从中选择两个数,使得(c_a^b)最大
猜结论:a必然是最大值,b就选择最接近a的一半的那个值即可
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef long long LL;
int n, a[N];
int main() {
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a, a + n);
LL res = 0;
for (int i = 0; i < n - 1; i++) {
if (abs(a[n - 1] - 2 * a[i]) < abs(a[n - 1] - 2 * res)) res = a[i];
}
cout << a[n - 1] << ' ' << res << endl;
return 0;
}