//题目1003:A+B #include<iostream> #include<string.h> using namespace std; int main() { int n1,n2; int s1[12],s2[12]; int s[12]; char c1[20],c2[20]; while(cin>>c1){ n1=0,n2=0;
1003:Hangover 查看提交统计提示提问 总时间限制: 1000ms 内存限制: 65536kB 描述 How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a ca
1003. Emergency (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue As an emergency rescue team leader of a city, you are given a special map of y
棋盘占领 Accepts: 937 Submissions: 2201 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description 百小度最近迷恋上了一款游戏,游戏里有一个n*m的棋盘,每个
题目连接:1003 Max Sum 题目大意:在一个数组中, 找到连续最大的和, 并输出它们的范围, 尽量长。 解题思路:累计求和, 复杂度o(n), 听别人讲了一下这种方法, 就来试试看了。 #include <stdio.h>const int N = 100005;const int low = -0xfffffff;int main() {int cas,
杭电题1003 Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 +
最大子序列和:HDU 1003 Max Sum 用动态规划的思想,求解前i个数能获得的最大和,当sum<0的时候,再加上a[i]就不如直接要a[i](不管a[i]是正是负),所以if sum < 0 then sum = 0,from = i+1;这里要更新起点 如果sum > ans则更新ans和起点、终点。 //Must so#include<iostream>#include<
1003. Emergency (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue As an emergency rescue team leader of a city, you are given a special map of y