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[Codeforces Round 636 (Div. 3)] E. Weights Distributing (BFS+最短路+贪心) E. Weights Distributing time limit per test 2 seconds memory limit per test 256 m 阅读全文
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[AtCoder Beginner Contest 163] F path pass i (树型dfs,容斥定律) 链接:https://atcoder.jp/contests/abc163/tasks/abc163_f Problem Statement We have a tree with N 阅读全文
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AtCoder Beginner Contest 163 E Active Infants (DP) Problem Statement There are NN children standing in a line from left to right. The activeness of th 阅读全文
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``` cpp const int maxn=255; int val[maxn][maxn]; const int log_maxn=8; int dpmin[maxn][maxn][log_maxn][log_maxn]; int dpmax[maxn][maxn][log_maxn][log_maxn]; void initRMQ(int n,int m) { mm[0]=-1; for(i 阅读全文
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牛客算法周周练2 链接:https://ac.nowcoder.com/acm/contest/5203/B 来源:牛客网 Music Problem 时间限制:C/C++ 2秒,其他语言4秒 空间限制:C/C++ 131072K,其他语言262144K 64bit IO Format: %lld 阅读全文
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牛客练习赛61 E 相似的子串(hash+二分) 链接:https://ac.nowcoder.com/acm/contest/5026/E 来源:牛客网 题目描述 给定一个字符串,要求取出k个位置不相交的子串,且他们之间任意两个的最长公共前缀的长度均不小于x。现在给出k,求最大的x。 两个 阅读全文
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AtCoder Beginner Contest 151 F Enclose All (最小圆覆盖) Problem Statement Given are NN points (xi,yi)(xi,yi) in a two dimensional plane. Find the minimum r 阅读全文
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[Codeforces Round 320 (Div. 2) E. Weakness and Poorness (三分/二分) E. Weakness and Poorness time limit per test 2 seconds memory limit per test 256 megab 阅读全文
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[Codeforces Round 320 (Div. 2) C. A Problem about Polyline (数学) C. A Problem about Polyline time limit per test 1 second memory limit per test 256 meg 阅读全文
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[Codeforces Round 630 (Div. 2)] E. Height All the Same (组合数学) E. Height All the Same time limit per test 2 seconds memory limit per test 512 megabytes 阅读全文