Codeforces Round #266 (Div. 2)_html/css_WEB-ITnose

Codeforces Round #266 (Div. 2)

Question link

A: Just judge which one is bigger

B: Enumerate x to sqrt( n), then you can directly calculate y, and then make a judgment

C: First judge whether the sum is a multiple of 3, and then preprocess the position of the prefix sum and the suffix sum corresponding to sum / 3 numbers, and then Scan from beginning to end and combine the current one with the next one

D: First preprocess the difference so that the array represents the way to add line segments, and then each time there is a -1, it can be combined with the previous one 1 to match, multiply the number of solutions by that number. If it is 0, it can match the previous one

E: Use union lookup to split each query into 2 parts, starting from x, x to the root, and then dfs from the root down each time. If the corresponding query is consistent, the corresponding query will be asked. After dfs, if a query is consistent twice, it will output YES if it is consistent, otherwise it will be NO

Code:

#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>