Problem Description
As a cheap labor in a small company, LL has to ride back and forth between school and office every day. It is a tedious trip. So he want to design the most satisfactory riding route. After several day's experiment, he draw the simple map. It contains n*m areas. The school is in (0,0) while the office is in (n-1,m-1). He also record the scenery rank of each area A(i,j) and the time need to ride through each area B(i,j). ( For the start and end, A(0,0), B(0,0), A(n-1,m-1), B(n-1,m-1) are always 0. ) Now, LL defines the satisfactory degree of a round trip as follow:

                                   ∑{ A(i,j) | Area (i,j) is in the riding route (come or go). }

the satisfactory degree = ----------------------------------------------------------------------
∑{ B(i,j) | Area (i,j) is in the riding route (come or go). }

Attention: 1. LL doesn't want to make a detour. So, from school to office he only ride rightward or downward and from office to school only leftward or upward.
2. LL won't pass the same area in the whole round trip except the start and end.

Each test case begins with two integers n,m ( 3<=n,m<=30 ), which is the size of the map. Then n lines follow, each contains m integers A(i,j). Another n lines follow, each contains m integers B(i,j). 1 <= A(i,j),B(i,j) <= 100.

For each case, Output the maximal satisfactory degree he can get in a round trip.

Sample Input
3 3
0 1 2
3 4 5
6 7 0
0 7 6
5 4 3
2 1 0

Sample Output

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本文档是帮助初学者化简你的c语言代码 表 + 程序 考虑以下问题: 一个n*m的方格,每个方格中有0或1这两个数字,现在请你输出一个n*m的矩阵,flag[i][j]表示在原来的第i行第j列处有多少个1环绕在周围。
用C语言如何解决Halloween treats这个算法问题?
Problem DescriptionrnEvery year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets on that day, no matter how many children call on him, so it may happen that a child will get nothing if it is too late. To avoid conflicts, the children have decided they will put all sweets together and then divide them evenly among themselves. From last year's experience of Halloween they know how many sweets they get from each neighbour. Since they care more about justice than about the number of sweets they get, they want to select a subset of the neighbours to visit, so that in sharing every child receives the same number of sweets. They will not be satisfied if they have any sweets left which cannot be divided. rnrnYour job is to help the children and present a solution. rnrn rnrnInputrnThe input contains several test cases. rnThe first line of each test case contains two integers c and n (1 ≤ c ≤ n ≤ 100000), the number of children and the number of neighbours, respectively. The next line contains n space separated integers a1 , ... , an (1 ≤ ai ≤ 100000 ), where ai represents the number of sweets the children get if they visit neighbour i. rnrnThe last test case is followed by two zeros. rnrn rnrnOutputrnFor each test case output one line with the indices of the neighbours the children should select (here, index i corresponds to neighbour i who gives a total number of ai sweets). If there is no solution where each child gets at least one sweet, print "no sweets" instead. Note that if there are several solutions where each child gets at least one sweet, you may print any of them.rn rnrnSample Inputrn4 5rn1 2 3 7 5rn3 6rn7 11 2 5 13 17rn0 0rn rnrnSample Outputrn3 5rn2 3 4
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c 语言知识 c 语言知识 c 语言知识 c 语言知识 c 语言知识 c 语言知识 c 语言知识 c 语言知识 c 语言知识
有## 有关C语言for语句的一些小知识 for (i = 1;i&lt;=4;i++) { scanf("%d", &amp;x); if (x != result) { if (i == 4) { if (c == 0) { printf("三次回答错误,正确答案是...
文件的有关基本知识 1 什么是文件 文件有俩中类型,在程序设计中,主要用到两种文件: 程序文件: 包括源程序文件(后缀为.c),目标文件(后缀为.obj),可执行文件(后缀为.exe).这种文件的内容为程序代码。 数据文件: 文件的内容不是程序,而是供程序运行时读写的数据。 我们主要学习数据文件 文件(file)是程序设计中一个重要概念。所谓的“文件”一般指储存在外部介质上的数据的集合。 2 文件...
C编译器对结构空间缺省的分配,结构中的位段更改C编译器的缺省分配策略,C 语言中的段位操作等等。。。
有关C语言的知识---宏定义用法 1.#include ----------------------引用另一个文件,从编译器自带的系统头文件中查找引用 2.#include “”xxxx.h”-----------------------引用另一个文件,优先从当前文件目录下进行查找 3.#if-#else-#endif ------------------------宏定义中的条件编译指令
任务09_1 根据已知的数据结构和main函数,完成create、sort和visit函数,严格按照运行效果进行数据的输入和输出,数据结构如下: #include&lt;stdio.h&gt; #include&lt;stdlib.h&gt; typedef struct student { int no; char name[15]; int sco...
习题1-5 打折 一件衣服95元,若消费满300元,打八五折。输入购买衣服的件数,输出需要支付的金额(单位:元),保留两位小数。 #include&amp;lt;stdio.h&amp;gt; #include&amp;lt;math.h&amp;gt; int main() { int n; double a; scanf(&quot;%d&quot;,&amp;amp;n); a=n95.0; if(a&amp;lt;300) { printf(&quot;%.2lf\...
算法的概念  广义地说,为解决一个问题而采取的方法和步骤,就称为“算法”。 著名计算机科学家沃思提出一个公式:      数据结构+ 算法= 程序 完整的程序设计应该是: 数据结构+算法+程序设计方法+语言工具 算法的表示 可以用不同的方法表示算法,常用的有: –自然语言 –传统流程图 –结构化流程图 –伪代码 –PAD图 模块设计的方法: •模
下面这个程序使用C语言的结构体实现了分数的加减乘除四则运算,同时将结果最简化#include <stdio.h>struct fraction { int numerator; int denominator; };// 将分数化为最简形式 struct fraction reduce_fraction(struct fraction f);// 分数的四则运算 struct fra
问题描述: 某侦察队接到一项紧急任务要求在A、B、C、D、E、F六个队员中尽可能多地挑若干人但有以下限制条件: 1)A和B两人中至少去一人; 2)A和D不能一起去; 3)A、E和F三人中要派两人去; 4)B和C都去或都不去; 5)C和D两人中去一个; 6)若D不去,则E也不去。 问应当让哪几个人去? 输入 输出 要派出的人 若有多个,按字母递增顺序排列,用逗
#include<stdio.h>int yue(int a, int b) { if (b != 0) yue(b, b%a); else return a; } int main() { int fz, fm; while (scanf("%d/%d",&fz,&fm) != EOF) { int a = yue(fz, fm);
真分数化简为最简分数 注意: 真分数(分母一定大于分子) #include&lt;iostream&gt; #include&lt;stdio.h&gt; using namespace std; //namespace是指标识符的各种可见范围 int main(){ int n;//分子 int m;//分母 int f...
c语言算法c语言 算法
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Problem Descriptionrn"Man, this year has the worst weather ever!", David said as he sat crouched in the small cave where we had sought shelter from yet another sudden rainstorm. rn"Nuh-uh!", Diana immediately replied in her traditional know-it-all manner. rn"Is too!", David countered cunningly. Terrific. Not only were we stuck in this cave, now we would have to listen to those two nagging for at least an hour. It was time to cut this discussion short. rn"Big nuh-uh. In fact, 93 years ago it had already rained five times as much by this time of year." rn"Duh", David capitulated, "so it's the worst weather in 93 years then." rn"Nuh-uh, this is actually the worst weather in 23 years.", Diana again broke in. rn"Yeah, well, whatever", David sighed, "Who cares anyway?". rnWell, dear contestants, you care, don't you? rnYour task is to, given information about the amount of rain during different years in the history of the universe, and a series of statements in the form "Year X had the most rain since year Y", determine whether these are true, might be true, or are false. We say that such a statement is true if: rnrnThe amount of rain during these two years and all years between them is known. rnrnIt rained at most as much during year X as it did during year Y. rnrnFor every year Z satisfying Y < Z < X, the amount of rain during year Z was less than the amount of rain during year X.rnrnWe say that such a statement might be true if there is an assignment of amounts of rain to years for which there is no information, such that the statement becomes true. We say that the statement is false otherwise.rn rnrnInputrnThe input will consist of several test cases, each consisting of two parts. rnThe first part begins with an integer 1 <= n <= 50000, indicating the number of different years for which there is information. Next follow n lines. The ith of these contains two integers -109 <= yi <= 109 and 1 <= ri <= 109 indicating that there was ri millilitres of rain during year yi (note that the amount of rain during a year can be any nonnegative integer, the limitation on ri is just a limitation on the input). You may assume that yi < yi+1 for 1 <= i < n. rnThe second part of a test case starts with an integer 1 <= m <= 10000, indicating the number of queries to process. The following m lines each contain two integers -109 <= Y < X <= 109 indicating two years. rnThere is a blank line between test cases. The input is terminated by a case where n = 0 and m = 0. This case should not be processed. rnTechnical note: Due to the size of the input, the use of cin/cout in C++ might be too slow in this problem. Use scanf/printf instead. In Java, make sure that both input and output is buffered.rn rnrnOutputrnThere should be m lines of output for each test case, corresponding to the m queries. Queries should be answered with "true" if the statement is true, "maybe" if the statement might be true, and "false" if the statement is false. rnSeparate the output of two different test cases by a blank line.rn rnrnSample Inputrn4rn2002 4920rn2003 5901rn2004 2832rn2005 3890rn2rn2002 2005rn2003 2005rnrn3rn1985 5782rn1995 3048rn2005 4890rn2rn1985 2005rn2005 2015rnrn0rn0rn rnrnSample Outputrnfalserntruernrnmaybernmaybe
Problem DescriptionrnMike has just found a dungeon. Inside there are n rooms and p treasure chests. Chests 1,...,p are located in room v1,...,vp respectively. The rooms are connected by n−1 passages. Any two rooms connected by a passage are considered adjacent. This dungeon is very special that there exists a path between any two rooms. Among these treasure chests, only chest 1 can be taken in the very beginning, others must be taken in order. That is, only after chest i is taken, may chest i+1 be taken as well, for every i∈1,...,p−1.rnrnMike is lazy, so he dislikes to explore the dungeon on his own feet. Therefore, Mike sends his robot dog, Blot, to collect the treasures in the dungeon. Blot can automatically collect the treasures from the chests, and it takes Blot exactly a second to move to any adjacent room. However, Blot’s movement is uncontrollable. Blot moves to all adjacent rooms with the same probability. For example, if there are k adjacent rooms, then Blot will move to any of them with probability 1k. Please help mike to compute the expected time that Blot collects all p treasures for him.rn rnrnInputrnThe first line on the input contains an integer T(T≤10). There will be T test cases. The first line of each test case contains an integer n(2≤n≤50000) indicating the number of rooms in the dungeon. The rooms are numbered from 0 to n−1. Each of the following n−1 lines contains two integers u and v(u,v∈0,...,n−1) indicating room u and room v are adjacent.rnrnThe next line contains an integer q(q≤100) indicating the number of scenarios. Each of the following q lines represents a scenario which consists of some integers p,v0,v1,...,vp separated by blanks. The first integer p(0
178个与算法有关的C语言源码 178个与算法有关的C语言源码,经典经典啊,还有一些计算代码,比如求两数的最大公约数、素数的求法、判断longint范围内的数是否为素数(包含求50000以内的素数表)、寻找离生成树最近的未加入顶点k、按权值递增顺序删去图中的边,若不形成回路则将此边加入最小生成树、计算图的传递闭包、无向图的连通分量、拓扑排序,找入度为0的点,删去与其相连的所有边,不断重复这一过程,例寻找一数列,其中任意连续p项之和为正,任意q 项之和为负,若不存在则输出NO等等,无意在CSDN发现的,有不少人在下载,发在这里,希望对C语言的学习有帮助。
![图片说明](https://img-ask.csdn.net/upload/201811/22/1542861843_263323.jpg) rnrn```rn#include rn#includernvoid main()rnrn int s=1;rn double n=1,t=1,pi=0;rn while(fabs(t)>1e-7)rn rn pi+=t;rn n+=2;rn s=-s;rn t=s/n;rn rn pi*=4;rn printf("%g",pi);rnrnrn rn```rnrnrn```rn #include rn#include rnrnint main()rnrn int i;rn double sum=0,pi,t; rn while(fabs(t)<10e-7)rn rn if(i%2==0)rn t=-1/(2*i-1);rn elsern t=1/(2*i-1);rn sum+=t;rnrn rn printf("%lf",sum*4.0000);rn return 0;rnrn```rnrn请问这两种方法为何输出结果不同
关于动态规划的资料 c/c++语言的有关动态规划的算法资料,可能对你有帮助
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