02 2015 档案
摘要://CodeForces 404B 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 const __int64 N = 1e6...
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摘要://CodeForces 404A 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 char mat[310][310]; 7...
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摘要://CodeForces 405B 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 char str[3010]; 7 int...
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摘要://CodeForces 405A 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 int ans[110], n; 7 8...
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摘要://CodeForces 407B 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 const __int64 mod = 1...
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摘要://CodeForces 407A 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 int square[1010]; 7 s...
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摘要://CodeForces 408B 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 int n1[30], n2[30]; 7...
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摘要://CodeForces 408A 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 int n, k[110], tmp, r...
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摘要://CodeForces 413C 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 413B 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 413A 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 412E 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 c...
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摘要://CodeForces 412D 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 412C 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 v...
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摘要://CodeForces 412B 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 412A 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 411C 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 411B 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 i...
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摘要://CodeForces 411A 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "vector" 6 using namespace std; 7 c...
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摘要://GCIS 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 int dp[510], Max; 7 int s1[510],...
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摘要://先贴一发错误的代码,dp[层数][方向]//因为是对不完整的方案计数了... 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; ...
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摘要://搜啊搜...//因为我们每次搜索的起点都是一样的,但是数组的第一个元素不同,所以可以手工算出下一个点,从起点的下一点开始搜 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" ...
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摘要://继续水一道树形dp 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "cmath" 6 using namespace std; 7 __int64 ...
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摘要://再水一发树形dp 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 int dp[6010][2]; 7 bool vis[...
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摘要://好久没水题了,水一发记忆化好了= ̄ω ̄= 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "cmath" 6 using namespace std;...
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摘要://思路:通过博弈的思想设计出状态,dp[w][b] 表示当公主面对 w 只白鼠和 b 只黑鼠时获胜的概率 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using na...
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摘要://分析:明显的树形关系,题目描述的是一棵高 n + 1的完全二叉树,则 dp[树层号][team号](规定最底层为 0 层,层数朝节点的方向依次递增),推一下就好了//稍微需要想一下的是比赛双方的选取,下面给出两种方法//#1 枚举起点划分team区间 1 #include "iostream" ...
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摘要://dp[i][state]//状态转移方程的选取:题目的意思是,从点 i 优先考虑到达 i + A 点的情况,若不能到达 i + A 点,则考虑到达 i + A + 1 点的情况...以此类推,直到考虑完到达 i + B 点的情况。对于最坏的情况,每一个点都有 B - A + 1 个后继点,所以由...
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摘要://下面是错误代码... 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 #include "cmath" 6 using namespace std; 7 const d...
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摘要://poj 2151 概率dp 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 double dp[33][33]; 7 i...
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摘要://CodeForces 385C素数朴素筛法,穷尽数的素数因子 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 const ...
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摘要://CodeForces 234F 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 const int INF = 0x3f3...
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摘要://CodeForces 232B//原先 T 了的代码 1 #include "iostream" 2 #include "cstdio" 3 #include "cstring" 4 #include "algorithm" 5 using namespace std; 6 const int ...
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摘要://CodeForces314C//分析:相当于求给定序列的不降子序列的个数,从一个空序列开始将得到的不降子序列不断的延长是典型的做法,则dp[i]表示以第 i 个元素结尾的序列//思路:O(n^2) 的做法,dp[i] = sum(dp[j]]) (a[j] = 1; i -= lowbit(i)...
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摘要://CodeForce18D - Seller Bob 1 #include"iostream" 2 #include"cstdio" 3 #include"cstring" 4 #include"algorithm" 5 using namespace std; 6 int num[2010][7...
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摘要://CodeForces 132C之前 wa 在 test 16 上的写法今天突发奇想改了改竟然过了......也就是说我这么找的状态也是可以的?!!! 1 #include"iostream" 2 #include"cstdio" 3 #include"cstring" 4 #include"al...
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摘要://CodeForces 264B//分析:在数列上的 dp,选取前缀集合的大小作为下标很容易构造出一个 O(n^2) 的 dp,i > j,if(ans[i]与ans[j]之间具有相同的素因子) dp[i] = max(dp[j]+1),不过 n 有100000那么大,于是继续优化,dp[i]表示...
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摘要://CodeForces 132C 1 #include"iostream" 2 #include"cstdio" 3 #include"cstring" 4 #include"algorithm" 5 using namespace std; //状态可达dp,其实 bool dp[110]...
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摘要://CodeChef RRMATRIX//分析:(x-1)*m+y == (y-1)*n+x ==> (x-1)*(m-1) == (y-1)*(n-1) ==> (x-1) == (y-1)*(n-1)/(m-1);(n-1)/(m-1)约分后分母还剩余一个(m-1)/gcd(n-1,m-1),(...
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摘要://poj 3666//分析:只是在2005年集训队论文黄源河提到的题目上略微有一点点变化 1 #include"iostream" 2 #include"cstdio" 3 using namespace std; 4 const int maxn = 2100; 5 int v[maxn],l[...
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摘要://poj 3016 K-Monotonic//分析:与2005年集训队论文黄源河提到的题目类似,给定序列a,求一序列b,b不减,且sigma(abs(ai-bi))最小。//思路:去除左偏树(大根堆)一半的节点(向上取整),让左偏树的根节点上存放中位数;每个左偏树的根节点表示一个等值区间//在本题...
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