Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. The cows are now tired of often being forced to take a particular path and want to build some new paths so that they will always have a choice of at least two...
Longge is good at mathematics and he likes to think about hard mathematical problems which will be solved by some graceful algorithms. Now a problem comes: Given an integer N(1 < N < 2^31),you are to calculate ∑gcd(i, N) 1<=i <=N. "Oh, I know, I know!" Longge shouts! But...
昵称:Bingsen 园龄:8年5个月 粉丝:0 关注:0 +加关注 评论排行榜 1. Educational Codeforces Round 22 补题 CF 813 A-F(2) 2. CODEFORCES 125E MST Company 巧用Kruskal算法(2) 3. 使用GAC加速 解决CSP问题 Kakuro - Cross Sums(1) 4. 第七周 Leetcode 466. Count The Repetitions 倍增DP (HARD)...
{ intnow=q.front();q.pop(); for(inti=0;i<connect[now].size();i++) { intnext=connect[now][i]; if(color[next]==0) { if(color[now]==1)color[next]=2; elsecolor[next]=1; q.push(next); continue; } if(color[next]==color[now])returnfalse; } } returntrue; } intmain()...
}if(id[now]==0) id[now]=++cnt;returnid[now]; } }tr;intdegree[25000*20];intF[25000*20];intfind(intx){if(F[x]==-1)returnx;returnF[x]=find(F[x]); }voidbing(inta,intb){intt1=find(a);intt2=find(b);if(t1!=t2) F[t1]=t2; ...
Bingsen 博客园 首页 新随笔 联系 订阅 管理 随笔- 128, 文章 - 0, 评论 - 6, 阅读 - 56086 POJ3420 递推+矩阵快速幂 POJ3420 很有趣的覆盖问题递归推导如下:f[n] = f[n-1] + 4*f[n-2] + 2 * [ f[n-3] + f[n-5] + f[n-7] +... ] + 3 * [ f[n-4] + f[n-6] +...
}intquery(intl,intr,intrt,intnowl,intnowr){if(nowl <= l && r <= nowr)returnw[rt]; push_col(l,r,rt);intm = (l+r)>>1;longlongans =0;if(nowl <= m) ans |= query(lson,nowl,nowr);//左边与右边要合并if(m < nowr)ans |=query(rson,nowl,nowr);returnans; ...
昵称:Bingsen 园龄:8年5个月 粉丝:0 关注:0 +加关注 评论排行榜 1. Educational Codeforces Round 22 补题 CF 813 A-F(2) 2. CODEFORCES 125E MST Company 巧用Kruskal算法(2) 3. 使用GAC加速 解决CSP问题 Kakuro - Cross Sums(1) 4. 第七周 Leetcode 466. Count The Repetitions 倍增DP (HARD)...
elserigh=nowans; } printf("%.3lf\n",(lef+righ)/2); } return0; } 用二分答案思想解决的生成树问题还有单度限制最小生成树,参考CODEFORCES 125E. 分类:Graph 好文要顶关注我收藏该文微信分享 Bingsen 粉丝-0关注 -0 +加关注 0 0 升级成为会员 ...