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"题目链接" 记$dp(i)$为以第i个结尾的上升子序列的个数，可以得到转移方程 $$dp(i)=\sum_{j=1,\,a[j] include include include include include include include include include include incl 阅读全文
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"题目链接" 模板题 按顺序BFS一边就可以判断结果了 C++ include include include include include include include include include include include include include using namespac 阅读全文
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"题目链接" 模板题 区间更新单点查询的树状数组 实现区间[i,j]增加val。 查询方式即前n项和 C++ include include include include include include include include include include include include 阅读全文
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"题目链接" 超级麻烦的一题，开始还理解错题目意思了。。。!看着udebug的数据才明白，重读一遍题目。。。 三个机器人移动一致，碰到障碍物的机器人就无法向障碍物方向移动，机器人自己也算障碍物。麻烦在机器人当作障碍物的地方。 标记状态BFS暴力 C++ include include include 阅读全文
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" Codeforces Round 429 C 题目链接 " 这题规律很好找，用最大的跟最小的匹配就好了，但是证明还是挺烦人的。听说是高考题。。。 $F(n,r) = 1p_1+2p_2+3p_3+\cdots+kp_k$, 其中k是选取的最小值，$p_k$表示的是选取最小值的概率。 $$F(n, 阅读全文
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众所周知，bellman_ford算法最常用的性质是判负环 也可以求最短路，SPFA算法改自Bellman_Ford。 1074 Extended Traffic $(u,v)$的边权是$ (b[v] b[u])^3$，输出从顶点1到各点最短路径，因为存在负环，所以从负环到达的点的最短路是 inf。 阅读全文
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"题目链接" 暴力枚举结果情况，然后跟当前比较，找到最小次数 C++ include include include using namespace std; int t; char s[1005]; int main() { scanf("%d", &t); while(t ) { scanf(" 阅读全文