Codeforces Round #266 (Div. 2)
题目链接
A:就简单的判断一下那种更大即可
B:枚举x到sqrt(n),然后可以直接算出y,然后判断一下即可
C:先判断和是否是3的倍数,然后预处理出前缀和出现位置和后缀和对应sum / 3个数,然后从头往后扫一遍把当前一个和后面进行组合即可
D:先预处理出差分,使得数组表示线段的添加方式,然后每次有一个-1,就能和前面多少个1进行匹配,方案数就乘上多少,如果是0,就能和前面+1个匹配
E:利用并查集,把每次询问拆分成2个部分,起点到x,x到根,然后每次从根往下dfs一遍,对应询问符合的就把对应询问++,dfs完如果一个询问符合两次,就是符合的输出YES,否则就是NO
代码:
#include <cstdio>#include <cstring>int n, m, a, b;int solve() { if (b >= m * a) return a * n; int yu = n % m; int ans = n / m * b; if (yu * a <br> B: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;ll n, a, b;int main() { scanf("%lld%lld%lld", &n, &a, &b); n = n * 6; ll ans = 1e18, x, y; if (a * b >= n) { x = a; y = b; ans = a * b; } else { int flag = 0; if (a > b) { flag = 1; swap(a, b); } for (int i = 1; i r) swap(l, r); if (l <br> C: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>const int N = 500005;typedef long long ll;int n;ll a[N], pres[N], prec[N], sufs[N], sufc[N];int main() { scanf("%d", &n); ll sum = 0; for (int i = 1; i = 1; i--) { sufs[i] = sufs[i + 1] + a[i]; sufc[i] = sufc[i + 1]; if (sufs[i] == sum) sufc[i]++; } ll ans = 0; for (int i = 1; i <br> D: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>typedef long long ll;const int MOD = 1000000007;const int N = 2005;int n, h, a[N], b[N];int main() { scanf("%d%d", &n, &h); for (int i = 1; i <br> E: <p></p> <p> </p> <pre name="code" class="sycode">#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;#define MP(a,b) make_pair(a,b)typedef pair<int int> pii;const int N = 100005;int n, m, parent[N];int find(int x) { return x == parent[x] ? x : parent[x] = find(parent[x]);}vector<pii> p, q[N];vector<int> g[N];int tot, vis[N], cnt[N];void dfs(int u) { vis[u] = 1; for (int i = 0; i <br> <br> <p></p> </int></pii></int></algorithm></vector></cstring></cstdio>