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今日已更新 2220 篇代码解决方案
题目真是不好读，大意如下（知道题意就很好解了）大致题意：一条河长度为 L，河的起点(Start)和终点(End)分别有2块石头，S到E的距离就是L。河中有n块石头，每块石头
https://www.u72.net/daima/fddz.html - 2024-07-09 19:29:29 - 代码库#include<stdio.h>int num[]={3,5,7,9,10,10,10,12,15,20,25,30};//10个数int lower_bound(int *A,int L,int R,int V){ int m; while(L<R) {
https://www.u72.net/daima/rm87.html - 2024-07-12 16:06:05 - 代码库给定n m kn和m为一个矩阵的行和列，都从1开始，矩阵的每个元素的值即为 i*j（行*列），求里面第k个数还想找什么规律，发现虽然矩阵里面很有规律，但是n 和m在不断变
https://www.u72.net/daima/sn23.html - 2024-07-12 17:39:34 - 代码库Problem HHalumTime Limit : 3 seconds You are given a directed graph G(V,E) with a set of vertices and edges. Each edge (i,j) that connects
https://www.u72.net/daima/w5ne.html - 2024-07-16 10:41:20 - 代码库给定NN个（长整型范围内的）整数，要求输出从小到大排序后的结果。本题旨在测试各种不同的排序算法在各种数据情况下的表现。各组测试数据特点如下： 数据1
https://www.u72.net/daima/w37f.html - 2024-08-26 00:01:12 - 代码库题设：设数组a[0],&hellip;,a[n-1]中存放着n>=1个已由小到大排序的不同整数，判断整数x是否在数组a中。若是，返回j，使得x=a[j];否则，返回-1。题解：利用数组中
https://www.u72.net/daima/s43n.html - 2024-07-13 10:24:26 - 代码库这是一个用户表，下面查询一个用户的数量：--查询表的用户数量SELECT COUNT(*) FROM UserInfo 再查询一下有几个男用户和几个女用户: SELECT C
https://www.u72.net/daima/sv1s.html - 2024-08-20 12:01:26 - 代码库FlyerTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1537 Accepted Submission(s)
https://www.u72.net/daima/wenb.html - 2024-07-16 15:10:58 - 代码库Problem HHalumTime Limit : 3 seconds You are given a directed graph G(V,E) with a set of vertices and edges. Each edge (i,j) that connects
https://www.u72.net/daima/w8zd.html - 2024-07-16 13:30:06 - 代码库Machine ScheduleTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8404 Accepted S
https://www.u72.net/daima/s8r4.html - 2024-08-21 04:18:43 - 代码库题意：输入一个地图，*是草坪 # 是墙 o是空地。机器人只能放在空地上，且两个同行或者同列的机器人中间必须有墙，问最多可以放多少个机器人。书上给出：....
https://www.u72.net/daima/veca.html - 2024-07-15 15:51:29 - 代码库# include <stdio.h># include <string.h># include <algorithm>using namespace std;int vis[200010];int b[200100];# define Max 200010int
https://www.u72.net/daima/warx.html - 2024-07-15 17:37:15 - 代码库直接搞。 想麻烦 也可以伞兵那样搞 ， 最小割网络流，把权值全置为1. 估计会超时。。#include <cstdio>#include <cstring>#include <algorithm>#include
https://www.u72.net/daima/vmu6.html - 2024-07-15 16:51:49 - 代码库题目链接：http://acm.hdu.edu.cn/showproblem.pihp?pid=2899题目大意：找出满足F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)的x值。注意精确度
https://www.u72.net/daima/v666.html - 2024-07-15 12:53:12 - 代码库#include<iostream>using namespace std;int findIndex(char * v,char key,int index){ int m,left,right; left=0; right=index; while(1
https://www.u72.net/daima/c52k.html - 2024-08-18 00:38:55 - 代码库// n点中选c点放下c头牛 是的n-1段距离中的最短距离最大 ，求这个最大的最短距离//假设当前的最小值为x，如果判断出最小差值为x时可以放下C头牛，//就先让
https://www.u72.net/daima/u9d5.html - 2024-07-14 14:47:03 - 代码库原题http://poj.org/problem?id=2456Aggressive cowsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6001 Accepted: 2989DescriptionFar
https://www.u72.net/daima/u9xd.html - 2024-07-14 15:02:23 - 代码库#include<string.h>#define MAXN 310#define inf 1000000000 #define _clr(x) memset(x,-1,sizeof(int)*MAXN)int KM(int m,int n,int mat[][MAXN
https://www.u72.net/daima/r39f.html - 2024-08-19 04:52:36 - 代码库http://acm.xidian.edu.cn/problem.php?id=1123 简直智障，scanf的格式都写错，T了无数遍。 #include<iostream>#include<cstring>#include<cstdio>
https://www.u72.net/daima/r44x.html - 2024-08-19 06:06:41 - 代码库//最大值最小//天数的a[i]值是固定的 不能改变顺序# include <algorithm># include <string.h># include <stdio.h>using namespace std;int n,m
https://www.u72.net/daima/vcw4.html - 2024-07-15 01:03:04 - 代码库