some code of HDOJ of mine
some code of HDOJ of mine
杭电OJ(1000-1099) AC 代码
杭州电子科技大学的离线版OJ,可用于断网情况下练习ACM
codj,hdoj的源码(50-60题)
杭电oj4405,一道简单的概率dp题目
标签: C++
HDOJ_my_answer
HDOJ 源代码 包含几百道HDOJ题目源码
标签: Java
HDACM1042import java.math.BigInteger; import java.util.Scanner;public class Main{ public static void main(String[] args) { Scanner sc = new Scanner(System.in); while(sc.hasNext()){
杭电acm解题报告 详细解析2000-2099 适合acm初学者
题目说明待求阶乘的数最大为10000,而10000!的位数为35660(这个数是上网查的),所以已经有的数据类型无法表示. 思路:用int型数组存储n!的每一步计算结果,并且数组大小应该不小于35660这个数....
HDOJ_1004 Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges’ favorite time is guessing the most popular problem. When the...
Problem Description You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of th
该题求解在给定的时间及条件下,判断是否存在一条从入口到出口的路。
#include<stdio.h> #include<string.h> int main(){ int i,n,len; scanf("%d",&n); char str[100]; getchar();...
#include<stdio.h> #include<math.h> int main() { int a,m,n,i,j,k,s,flag,time; while(scanf("%d %d",&m,&n)!=EOF){ ... for(a=m,fla...
#include<stdio.h> int main() { int n; while(scanf("%d",&n)!=EOF){ int a; int sum=1; //跳出一个循环免得值重复 while(n--){ scanf("%d"... ...
#include<stdio.h> #include<math.h> int main() { double x1,y1,x2,y2; while(scanf("%lf %lf %lf %lf",&x1,&...,sqrt((x2-x1)*(x2-x1)...
#include<iostream> #include<string.h> using namespace std; int main(){ int n,count; char str[1000]; scanf("%d",&...
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1010 题目大意:给出一个迷宫(含起点和终点),要求找出一条路径,这条路径的长度必须为某个规定的长度. ... ...在理解迷宫问题的基础上,再做本题....
标签: 水题
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2025 //C++代码 #include #include #include using namespace std; int main(){ char a[105],ch; while(cin.getline(a,105)){ ... ch=a
題目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1003题目大意,给定一个数组,a[0], a[1], ..., a[n-1]需要求它的一个连续子序列,使得这个连续子序列的和最大.一、暴力求解方法O(n^3)直观的解法是,遍历所有...
杭电oj2005, 题目: 给定一个日期,输出这个日期是该年的第几天。Input 输入数据有多组,每组占一行,数据格式为YYYY/MM/DD组成
N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 75189 Accepted Submission(s): 21920 Total Submission(s): 75189 Accepted Submis
#include #include #include #include #include #include using namespace std; int Start = 0; int End = 0; int maxSubSum(const vector& a) { int maxSum = -1, thisSum = 0;... for (int j =
HDOJ练习题目源码,约100多道题目,附带题目类型等等
标签: 水题
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2019 //C++代码 #include using namespace std; int main(){ int n,m,a,i; while(cin>>n>>m,n||m){ bool flag=0; for(i=1;i;...f
N! Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 64180 Accepted Submission(s): 18255 Problem Description ...Given an integer N(0
HDOJ_1480 钥匙计数之二 解题报告.mhtHDOJ_1480 钥匙计数之二 解题报告.mht
#include<iostream> #include<cstring> using namespace std; int main(){ char str[101]; while(gets(str)){ for(int i = 0;i <... if(i == 0||str[i-...
// hdoj_1089 A+B for Input-Output Practice (I) #include int main(void) { int a, b; while(scanf("%d%d", &a, &b) != EOF) { printf("%d\n", a + b); } return 0; } 水题,不解释……