Codeforces Round #268 (Div. 2)_html/css_WEB-ITnose

Codeforces Round #268 (Div. 2)

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A: Just mark some and judge them

B: Just judge the enumeration times one by one. Can

C: Construct it. 4 and 5 are constructed manually respectively. Then each time there are 2 more numbers, subtract them first to get 1, then multiply by the original number and it remains unchanged. Numbers below 4 cannot be constructed at all.

D: Greedy, sort first, and then each twopointer selects the first and last two to determine which set it can throw into. If it doesn’t work, just find a satisfactory one and throw it into the small set

E: For reasoning, see the official solution for details. After deducing, the interval [x, x 1e18 - 1] is moved one at a time to become [x 1, x 1e18], and the corresponding sum is increased by 1. So as long as it can be calculated [1, 1e18], and then move the corresponding number of steps to get the corresponding interval. Calculate the sum and push rules of 1-1e18 to find out. The official solution also has the formula

Code:

A:

#include <cstdio>#include <cstring>int n, p, q, vis[105];bool solve() {	for (int i = 1; i  <br> B: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>const int N = 55;int p, q, l, r, ans;int vis[10005], c[N], d[N];bool judge(int t) {	for (int i = 0; i  <br> C: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;int n;void solve(int n) {	if (n % 2 == 0) {		printf("1 + 2 = 3\n");		printf("3 + 3 = 6\n");		printf("6 * 4 = 24\n");		for (int i = 5; i  <br> D: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <map>#include <algorithm>using namespace std;const int N = 100005;int n, a, b;map<int int> to;struct Seq {	int num, id, to, vis;} s[N];bool cmp(Seq a, Seq b) {	return a.num  b || s[st].num + s[ed].num  b) {		swap(a, b);		flag = 1;	}	for (int i = 0; i  <br> E: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const ll INF = 1e18;ll a;int main() {	scanf("%lld", &a);	ll num = INF / 10 % a;	num = num * 2 % a;	num = num * 9 % a;	num = num * 9 % a;	num = num * 5 % a;	printf("%lld %lld\n", a - num, a - num + INF - 1);	return 0;}</algorithm></cstring></cstdio>