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;}
#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;}
#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;}
#include#include #include
#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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