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Given a collection of items s={1,2,3,...,n}, the weight of item I is WI, its value is VI, the capacity of the backpack is w, that is, the maximum load weight does not exceed W. Within a limited total weight of W, how do we select items to make the total value of the item the most. If the item can not be divided, that is, the item I is either the entire selection, or not selected, can not put the item I in the backpack multiple times, and can not only load part I, then the problem is called

Description Calculate the number of occurrences of 1 in the natural number 0-000000.The train of thought should consider six characters: 000000-999999, with a total of six characters * 1000 000 characters = characters. 0-9 is a probability of occurrence, so the number of 0-9 is 6 000 000/

Preface In Linux, all devices are regarded as files. Each time a file is opened, there is a file descriptor that represents the file to be opened. When the program starts, three I/O device files are opened by default: stdin, stdout, and stderr. The file descriptors 0, 1, and 2 are obtained respectively. Instance Now let's look at an example of testing the ttyname function (the ttyname function returns the p

Dynamic Programming 0-1 knapsack problem? Description: N items and a backpack are given. The Weight of item I is wi, its value is VI, and the size of the backpack is C. How should I select the items to be loaded into the backpack to maximize the total value of the items in the backpack ?? For an item, either it is packed into a backpack or it is not loaded. Therefore, the loading status of an item can be

#include 3 #include 4 long long f[10000]; 5 int n; 6 int main(){ 7 FILE *in,*out; 8 in=fopen("input6.txt","r"); 9 out=fopen("output.txt","w");10 fscanf(in,"%d",n); 11 if(((n*n+n)/2)%2==1)12 {13 fprintf(out,"0\n");14 printf("0\n");15 } 16 else 17 {18 int i,j;19 mem

set to 0 (that is, optimization is not required, __debug _ is set to true), there is no optimization. If the value is set to 1, the assert statement is deleted, __debug _ is set to false. If the value is set to 2, in addition to setting the value to 1, the instructions in the code will also be deleted to achieve the best optimization results. During code compila

The two days to learn the timer of the lpc2103. I started to look at the previous register introduction. I felt dizzy and I was confused later. However, when I see the following operations using the image to describe the timer-related registers, it will be clear. After you know the application, you can see the principle again. Why are these two timers put together, because they are both 32-bit timers, which are the same except for the peripheral base address. Let's talk about the features of th

Executive program Times wrong, Baidu search, first search this blog http://blog.csdn.net/huqiao1206/article/details/50768481, think the solution is too troublesome,Continue to search, and search this blog http://www.cnblogs.com/zhongtang/p/7448823.html (has been reproduced by me, on a blog is, afraid of the day his blog is not),Look at it is the solution, is ready to search again, suddenly found in the bottomThere is also a solution, (I assume you have read the previous blog), according to the s

1. Let's talk about the role of "# If 0/# If 1... # endif" first. You have mentioned a little bit above, but you have not mentioned any key points. We know that the C standard does not provide single-line style annotations such as "//" in C ++, but only provides block annotation functions such, we usually use it to write descriptive comments (comments) in the Cod

Exit seems to be in the "stdlib. H" header file, so it must contain the header fileReturn indicates a function call. If the main function is returned, the system exits the program.Exit is to forcibly exit the program at the call. Once the program is run, it will end. Exit (1) indicates an abnormal exit. This 1 is returned to the Operating SystemExit (0) indicates

1 /*2 depending on the relationship, you can build a tree3 Dp[i][j] Indicates the maximum value of the first node selected J4 Dp[i][j]=max (Sigma (Dp[c[i][ki]))5 Sigma (Dp[c[i][ki]]) indicates the maximum number of J points selected from the son node of I6 */7 /*#include 8 #include 9 #include Ten #include One using namespace std; A - const int maxn=205; - int DP[MAXN][MAXN]; the int VAL[MAXN],N,M,NUM[MAXN]; - vector - inline int max (int a,int b) {r

Tags: field query ISP Queued data Germany distributed Interview Database indexAll articles about database indexing will tell you not to Jianjian the word index. But the fact that these articles will not tell you is:The index is very useful if the value of this field in the table is extremely uneven, and you need to query for less distributed records.For example, if there are 10 million records in a table, and a record with a status of

First of all, I would like to thank my brother Xiaoxiao for helping you. Directly give the state transition equation: DP [I] [v] = max (DP [I-1] [v], DP [I-1] [V-Vol [I] + val [I]); Vol [I] is the volume of the I-th item, and Val is the value of the I-th item The equation means that when we consider the I items, there are only two options for I items: Put or not put, if not put the weight is the weight of