码代码的猿猿

摘要： 实质是：求一个字符串与其逆字符串的LCS.LCS用short int开数组刚好可以卡着内存过！！！ PalindromeTime Limit:3000MSMemory Limit:65536KTotal Submissions:46566Accepted:15910DescriptionA palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are... 阅读全文

摘要： The TriangleTime Limit: 1000MSMemory Limit: 10000KTotal Submissions: 32637Accepted: 19322Description73 88 1 02 7 4 44 5 2 6 5(Figure 1)Figure 1 shows a number triangle. Write a pro... 阅读全文

摘要： LIS：维护一个单调的队列新的元素，如果大于队尾元素，即插入队尾否则二分查找比它大的最小元素，替换掉最后队列长度即为LIS的解1 3 7 5 9 4 811 31 3 71 3 51 3 5 91 3 4 91 3 4 8 Longest Ordered SubsequenceTime Limit:2000MSMemory Limit:65536KTotal Submissions:26637Accepted:11611DescriptionA numeric sequence ofaiis ordered ifa1<a... 阅读全文

摘要： 1，按大小顺序帮最小2，dp[i][j] 前 i 件物品搬 j 对3，dp[i][j]=min(dp[i-1][j],dp[i-2][j-1]+POW2(a[i-1],a[i]);4，dp[i][i>>1]=dp[i-2][i>>1-1]+POW2(a[i-1],a[i]);搬寝室Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12181Accepted Submission(s): 4074Problem Descri 阅读全文

摘要： 计算出对应的高度，一层一层的DP最大面积City GameTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3179Accepted Submission(s): 1318Problem DescriptionBob is a strategy game programming specialist. In his new city building game the gaming environment is as follows: a cit 阅读全文

摘要： DP求矩形左右可以扩展的范围Largest Rectangle in a HistogramTime Limit: 1000MSMemory Limit: 65536KTotal Submissions: 12112Accepted: 3927DescriptionA histogram is a polygon composed of a sequence of rectangles align... 阅读全文

摘要： 最大连续子序列Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14568 Accepted Submission(s): 6363Problem Description给定K个整数的序列{ N1, N2, ..., NK }，其任意连... 阅读全文

摘要： A Different TaskThe (Three peg) Tower of Hanoi problem is a popular one in computer science. Briefly the problem is to transfer all the disks from peg-Ato peg-Cusing peg-Bas intermediate one in such a way that at no stage a larger disk is above a smaller disk. Normally, we want the minimum number of 阅读全文

摘要： 斯特林数S(p,k)的一个组合学解释是：将p个物体划分成k个非空的不可辨别的（可以理解为盒子没有编号）集合的方法数。S(p,k)的递推公式是：　S(p,k) = k*S(p-1,k) + S(p-1,k-1) ,1<= k <=p-1边界条件：S(p,p) = 1 ,p>=0S(p,0) = 0 ,p>=1一卡通大冒险Time Limit : 2000/1000ms (Java/Other)Memory Limit : 32768/32768K (Java/Other)Total Submission(s) : 5Accepted Submission(s) : 3Pr 阅读全文

摘要： 最大报销额Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12524Accepted Submission(s): 3452Problem Description现有一笔经费可以报销一定额度的发票。允许报销的发票类型包括买图书（A类）、文具（B类）、差旅（C类），要求每张发票的总额不得超过1000元，每张发票上，单项物品的价值不得超过600元。现请你编写程序，在给出的一堆发票中找出可以报销的、不超过给定额度的最大报销额。Input测试输入包含若 阅读全文