【CF比赛】Educational Codeforces Round 102 (Rated for Div. 2)
题目来源
Educational Codeforces Round 102 (Rated for Div. 2)
A. Replacing Elements
只要判断最大值是否小于等于d，或者第一个值加第二个值小于等于d即可。
#include<bits/stdc++.h>
using namespace std;#define _for(i, a, b) for (int i = (a); i < (b); ++i)
const int N = 110;
int a[N];int main()
{int T;scanf("%d", &T);while (T--){int n, d;scanf("%d%d", &n, &d);_for(i, 0, n) scanf("%d", &a[i]);sort(a, a + n);printf("%s\n", a[n - 1] <= d || a[0] + a[1] <= d ? "YES" : "NO");}return 0;
}
B. String LCM
思维题
如果两个字符串之前存在公共子串的话，那么他们长度的最大公约数也是也是一样的。
#include<bits/stdc++.h>
using namespace std;string a, b;int gcd(int a, int b)
{return b ? gcd(b, a % b) : a;
}string repeat(string& s, int n)
{string t;while (n--) t += s;return t;
}int main()
{int T;scanf("%d", &T);while (T--){cin >> a >> b;int d = gcd(a.size(), b.size());string c[] = { repeat(a, b.size() / d), repeat(b, a.size() / d) };cout << (c[0] == c[1] ? c[0] : "-1") << endl;}return 0;
}
C No More Inversions
这题转了怎么大个弯，就是要你求出在1~k的全排列中逆序数和a数组一样多且字典数最大的数组p，所以直接构造就好了，把数组a中k ~ n-(n - k)做p的后部分，a中剩下的没有重复的数字做p的前部分。
#include<bits/stdc++.h>
using namespace std;#define _for(i, a, b) for (int i = (a); i < (b); ++i)
#define _rep(i, a, b) for (int i = (a); i <= (b); ++i)int main()
{#ifdef LOCALfreopen("data.in", "r", stdin);
#endif int T;scanf("%d", &T);while (T--){int n, k;scanf("%d%d", &n, &k);_for (i, 1, 2 * k - n) printf("%d ", i);_rep(i, 0, n - k) printf("%d ", k - i);puts("");}return 0;
}
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