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
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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 <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;}
#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;}
#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;}
#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;}
#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;}
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