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Codeforces Round #268 (Div. 2)

题目链接

A：有的标记掉判断一下即可

B：枚举时间一个个去判断一下即可

C：构造一下，4和5分别手动构造，然后之后每次多2个数字就先相减掉得到1，然后乘上原数字不变，4以下的是绝对构造不出来的

D：贪心，先排序，然后每次twopointer选头尾两个判断能丢进哪个集合，都不行就找一个之前满足的丢进小集合

E：推理，详细见官方题解，大致是推出来后，[x, x + 1e18 - 1]这个区间，每次挪动一个变成[x + 1, x + 1e18]，对应的总和就加1，那么只要能求出[1, 1e18]的值，然后在去挪动相应步数得到相应区间即可，计算1-1e18的和推推规律就可以发现了，官方题解上也有公式

代码：

A：

#include #include 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;}

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B：

#include #include 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;}

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#include #include #include #include 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;}

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#include #include #include #include using namespace std;const int N = 100005;int n, a, b;map 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;}

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#include #include #include 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;}

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