UVa 10747 - Maximum Subsequence】的更多相关文章

题目大意:给出n个数,从中选取k个,使得乘积最大,并且尽量使和最大 分析:首先按照数的绝对值大小排序.然后就要分三大类情况讨论: (1)前k个中选到0:如果选到0的话,乘积一定是0,所以尽量选大的数,让和变大. (2)前k个中选到负数的个数为偶数:这样的话直接输出答案(一定为最优解) (3)前k个中选到的负数个数为奇数:这类情况比较复杂,还要分成两个子类: a)k个中没有正数: 如果换正数:优先用正数替换最小的负数:否则注定乘积为负数或者0:选最大的k个. b)k个中有正数: 还有正数和负数:比…
UVA - 11404 Palindromic Subsequence 题意:一个字符串,删去0个或多个字符,输出字典序最小且最长的回文字符串 不要求路径区间DP都可以做 然而要字典序最小 倒过来求LCS,转移同时维护f[i][j].s为当前状态字典序最小最优解 f[n][n].s的前半部分一定是回文串的前半部分(想想就行了) 当s的长度为奇时要多输出一个(因为这样长度+1,并且字典序保证最小(如axyzb  bzyxa,就是axb|||不全是回文串的原因是后半部分的字典序回文串可能不是最小,多…
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For examp…
1​​, N2N_2N​2​​, ..., NKN_KN​K​​ }. A continuous subsequence is defined to be { NiN_iN​i​​, Ni+1N_{i+1}N​i+1​​, ..., NjN_jN​j​​ } where 1≤i≤j≤K1 \le i \le j \le K1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum…
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For examp…
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For examp…
Given a sequence of K integers { N​1​​, N​2​​, ..., N​K​​ }. A continuous subsequence is defined to be { N​i​​, N​i+1​​, ..., N​j​​ } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For…
题目描述 HZ偶尔会拿些专业问题来忽悠那些非计算机专业的同学.今天测试组开完会后,他又发话了:在古老的一维模式识别中,常常需要计算连续子向量的最大和,当向量全为正数的时候,问题很好解决.但是,如果向量中包含负数,是否应该包含某个负数,并期望旁边的正数会弥补它呢?例如:{6,-3,-2,7,-15,1,2,2},连续子向量的最大和为8(从第0个开始,到第3个为止).你会不会被他忽悠住?     Given a sequence of K integers { N1, N2, ..., NK }.…
MSS(Array[],N)//Where N is the number of elements in array { sum=; //current sum max-sum=;//Maximum Sum seq-start=;//start of the subsequence seq-end=;//end of the subsequence ;i<N;i++){ sum=sum+Array[i]; ){ sum=; seq-start++; } else{ if(sum>max-sum…
Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For examp…