题意:给一段数列,将这个数列分成两部分,使两部分的最大子段和的和最大,输出和 /* 看数据没想到是(O)n的算法,求出从前向后的最大子段和和从后向前的最大子段和, 然后枚举断点. 第一次提交不小心折在数组最小值的赋值上-- */ #include<cstdio> #include<iostream> #include<cstring> #define M 50010 #define INF 1000000000 using namespace std; int a[M]…

ACdreamOJ 1154 Lowbit Sum (数位dp) ACM 题目地址:pid=1154" target="_blank" style="color:rgb(0,136,204); text-decoration:none">ACdreamOJ 1154 题意: long long ans = 0; for(int i = 1; i <= n; i ++) ans += lowbit(i) lowbit(i)的意思是将i转化成二进制…

[BZOJ3817/UOJ42]Sum(类欧) 题面 BZOJ UOJ 题解 令\(x=\sqrt r\),那么要求的式子是\[\sum_{d=1}^n(-1)^{[dx]}\] 不难发现,对于每个\(d\)而言的取值只和\([dx]\)的奇偶性相关. 如果\(x\)是个整数,也就是\(r\)是完全平方数的时候,显然是可以直接算答案的. 计算答案的时候显然之和有几个奇数或者几个偶数相关(只要求一个另外一个就是补集) 比如说我们来求有几个是偶数,那么要满足的条件就是\([dx]=2*[\frac{…

题目描述: Maximum Value time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a sequence a consisting of n integers. Find the maximum possible value of (integer remainder of *a**i* divi…

1146. Maximum Sum Time limit: 1.0 second Memory limit: 64 MB Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this…

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. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For exampl…

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to…

题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&page=show_problem&problem=44  Maximum Sum  Background A problem that is simple to solve in one dimension is often much more difficult to solve in more th…

Maximum Sum 大意:给你一个n*n的矩阵,求最大的子矩阵的和是多少. 思路:最開始我想的是预处理矩阵,遍历子矩阵的端点,发现复杂度是O(n^4).就不知道该怎么办了.问了一下,是压缩矩阵,转换成最大字段和的问题. 压缩行或者列都是能够的. int n, m, x, y, T, t; int Map[1010][1010]; int main() { while(~scanf("%d", &n)) { memset(Map, 0, sizeof(Map)); for(i…

Charm Bracelet    POJ 3624 就是一道典型的01背包问题: #include<iostream> #include<stdio.h> #include<algorithm> #include<string.h> using namespace std; ],b[]; ]; int main() { int n,m,i,j; while(scanf("%d%d",&n,&m)!=EOF) { ;i&l…