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 <= n; i++)		if (vis[i] == 0) return false;	return true;}int main() {	scanf("%d", &n);	scanf("%d", &p);	int tmp;	for (int i = 0; i < p; i++) {		scanf("%d", &tmp);		vis[tmp] = 1;	}	scanf("%d", &q);	for (int i = 0; i < q; i++) {		scanf("%d", &tmp);		vis[tmp] = 1;	}	printf("%s\n", solve() ? "I become the guy." : "Oh, my keyboard!");	return 0;}

B:

#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 < q; i++) {		for (int j = c[i]; j <= d[i]; j++) {			if (vis[j + t])				return true;		}	}	return false;}int main() {	scanf("%d%d%d%d", &p, &q, &l, &r);	int a, b;	for (int i = 0; i < p; i++) {		scanf("%d%d", &a, &b);		for (int j = a; j <= b; j++)			vis[j] = 1;	}	for (int i = 0; i < q; i++)		scanf("%d%d", &c[i], &d[i]);	for (int i = l; i <= r; i++) {		if (judge(i)) ans++;	}	printf("%d\n", ans);	return 0;}

C:

#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 <= n; i += 2) {			printf("%d - %d = 1\n", i + 1, i);			printf("24 * 1 = 24\n");		}	} else {		printf("5 - 3 = 2\n");		printf("1 + 2 = 3\n");		printf("2 * 3 = 6\n");		printf("4 * 6 = 24\n");		for (int i = 6; i <= n; i += 2) {			printf("%d - %d = 1\n", i + 1, i);			printf("24 * 1 = 24\n");		}	}}int main() {	scanf("%d", &n);	if (n < 4) printf("NO\n");	else {		printf("YES\n");		solve(n);	}	return 0;}

D:

#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.num;}bool cmpid(Seq a, Seq b) {	return a.id < b.id;}int flag = 0;bool solve() {	int st = 0, ed = n - 1;	while (st <= ed) {		if (s[st].vis) {			st++;			continue;		}		if (s[ed].vis) {			ed--;			continue;		}		if (s[st].num + s[ed].num > b || s[st].num + s[ed].num < a)			return false;		if (s[st].num + s[ed].num == b) {			s[st].vis = 1;			s[ed].vis = 1;			s[st].to = 1;			s[ed].to = 1;			st++;			ed--;			continue;		} else if (s[st].num + s[ed].num == a) {			s[st].vis = 0;			s[ed].vis = 0;			s[st].to = 0;			s[ed].to = 0;			st++;			ed--;		} else {			if (!to.count(a - s[st].num)) return false;			int v = to[a - s[st].num];			if (s[v].vis) return false;			s[v].to = 0;			s[v].vis = 1;			s[st].vis = 1;			s[st].to = 0;			st++;		}	}	sort(s, s + n, cmpid);	printf("YES\n");	printf("%d", s[0].to^flag);	for (int i = 1; i < n; i++)		printf(" %d", s[i].to^flag);	printf("\n");	return true;}int main() {	scanf("%d%d%d", &n, &a, &b);	if (a > b) {		swap(a, b);		flag = 1;	}	for (int i = 0; i < n; i++) {		scanf("%d", &s[i].num);		s[i].id = i;	}	sort(s, s + n, cmp);	for (int i = 0; i < n; i++)		to[s[i].num] = i;	if (!solve()) printf("NO\n");	return 0;}

E:

#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;}