看到了大神的代码.理解了好久...真是差距. 题意:给出m, p, a, b,然后xi满足已下两个公式, 求 xp1 + xp2 +...+ xpm 的最大值. 1.-1/sqrt(a) <= xi <= sqrt(a); (a>0) 2.x1+x2+...+xm = b*sqrt(a); 注意:p为偶数. 解题思路:因为p为偶数,所以sqrt(a)和-1/sqrt(a)的p次方都为正数且sqrt(a) > 1/sqrt(a).所以贪心思想时尽量先取sqrt(a);当已经取的xi的…

题目大意:UVa 108 - Maximum Sum的加强版,求最大子矩阵和,不过矩阵是可以循环的,矩阵到结尾时可以循环到开头.开始听纠结的,想着难道要分情况讨论吗?!就去网上搜,看到可以通过补全进行处理,也是,通过补全一个相同的,问题就迎刃而解了,所以把n*n的矩阵扩展成2n*2n的矩阵就好了. #include <cstdio> #include <cstring> #define MAXN 160 int a[MAXN][MAXN], sum[MAXN][MAXN]; int…

UVA.10305 Maximum Product (暴力) 题意分析 直接枚举起点和重点,然后算出来存到数组里面,sort然后取最大值即可. 代码总览 #include <iostream> #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <sstream> #include <set> #inc…

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…

题目传送门 /* 贪心:按照执行时间长的优先来排序 */ #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> #include <string> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; struct Node { int b, j; bo…

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F - Maximum GCD Time Limit:1000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Given the N integers, you have to nd the maximum GCD (greatest common divisor) of every possiblepair of these integers.InputThe rst line of input is an integer N (…

注意long long  long long  longlong !!!!!!   还有 printf的时候 明明longlong型的答案 用了%d  WA了也看不出,这个细节要注意!!! #include <cstdio> ]; int main() { ;long long ans,sum; while(~scanf("%d",&n)) { ;i<n;i++) scanf("%d",&a[i]); ans=; ;i<n;i…

Maximum Score Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=83008#problem/J Description Ron likes to play with integers. Recently he is interested in a game where some integers are given and he is…

题目链接:https://vjudge.net/contest/210334#problem/B 题目大意:Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of the maximum positive product involving consecutive terms of S. If you cannot find a positive sequen…