首页  >  文章  >  web前端  >  Codeforces Round #266 (Div. 2)_html/css_WEB-ITnose

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

WBOY
WBOY原创
2016-06-24 11:57:20983浏览

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>
声明:
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系admin@php.cn