hdoj分类】的更多相关文章

http://blog.csdn.net/lyy289065406/article/details/6642573 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128…
HDOJ 题目分类 //分类不是绝对的 //"*" 表示好题,需要多次回味 //"?"表示结论是正确的,但还停留在模块阶 段,需要理解,证明. //简单题看到就可以敲的 1000:    入门用: 1001:    用高斯求和公式要防溢出 1004:1012: 1013:    对9取余好了 1017:1021: 1027:    用STL中的next_permutation() 1029:1032:1037:1039:1040:1056:1064:1065: 10…
HDOJ 题目分类 /* * 一:简单题 */ 1000:    入门用:1001:    用高斯求和公式要防溢出1004:1012:1013:    对9取余好了1017:1021:1027:    用STL中的next_permutation()1029:1032:1037:1039:1040:1056:1064:1065:1076:    闰年 1084:1085:1089,1090,1091,1092,1093,1094, 1095, 1096:全是A+B1108:1157:1196:1…
one recursive approach to solve hdu 1016, list all permutations, solve N-Queens puzzle. reference: the video of stanford cs106b lecture 10 by Julie Zelenski https://www.youtube.com/watch?v=NdF1QDTRkck // hdu 1016, 795MS #include <cstdio> #include &l…
a typical variant of LCS algo. the key point here is, the dp[][] array contains enough message to determine the LCS, not only the length, but all of LCS candidate, we can backtrack to find all of LCS. for backtrack, one criteria is dp[i-1][j]==dp[i][…
reference: Rabin-Karp and Knuth-Morris-Pratt Algorithms By TheLlama– TopCoder Member https://www.topcoder.com/community/data-science/data-science-tutorials/introduction-to-string-searching-algorithms/ // to be improved #include <cstdio> #include <…
reference: 6.4 knapsack in Algorithms(算法概论), Sanjoy Dasgupta University of California, San Diego Christos Papadimitriou University of California at Berkeley Umesh Vazirani University of California at Berkeley the unbounded knapsack and 0-1 knapsack a…
three version are provided. disjoint set, linked list version with weighted-union heuristic, rooted tree version with rank by union and path compression, and a minor but substantial optimization for path compression version FindSet to avoid redundanc…
use fgets, and remove the potential '\n' in the string's last postion. (main point) remove redundancy there must be a stack, at first sight, you need a stack of type myNode, but think deeper, matrix multiplication is valid only if A.c=B.r, then the n…
the algorithm of three version below is essentially the same, namely, Kadane's algorithm, which is of O(n) complexity. https://en.wikipedia.org/wiki/Maximum_subarray_problem the evolution of the implementations is to remove redundancy and do what is…