c gpu programming tutorial

Problem Descriptionn individuals in a circle, numbered from 1 to n sequentially. From the person numbered 1 starts 1 to K, where the number of people who count K out of the circle, the output of the last left a person's original numbers.InputFirst enter a T, which indicates that there is a T group of data (1Then there are t-lines, each with 2 positive integers n and K. (1OutputFor each set of test data, output a number that represents the number of the last person left.Sample Input310 37 15 4Sam

Problem DescriptionEnter n (nInputExample of multi-group test sample. The first row of each group has an integer n indicating that there are n strings. Next there are n rows, one string per line.OutputOutput a sequential string, with each line outputting a string.Sample Input3abaaabcabSample OutputAababacab1#include 2#include string.h>3 voidSortChar* str[],intsize)4 {5 inti,j;6 Char*tmp;7 for(i=0; i1; i++)8 {9 for(j=i+1; j)Ten { One if(strcmp (str[i]

Problem DescriptionEnter n, and the output corresponds to a hollow positive hexagon with a side length of N.For easy viewing, the sample midpoint '. ' Represents a space and prints a space instead of a small dot when plotting a drawing.InputEdge length N. (NOutputSide length is a positive hexagon of nSample Input5Sample Output.....*****....*.....*...*.......*.. *.........*.*...........*.. *.........*...*.......*....*.....*.....*****1#include 2#include 3 voidPrtCharCintcount)4 { 5 while(cou

Problem DescriptionEnter n strings (nInputMultiple sets of test data. The first row of each set of test data contains an integer n, representing a total of n strings. Next, each line contains a string that consists of printable characters.OutputEach set of test sample output one row. The output finds a string that satisfies the test instructions.Sample Input3djdlkfjsadfjwedlkfjdlkfjl;jf;sfjdsl;al/dljfd2dlkasfjmmlld;femflsad;fiwejdifSample OutputDlkfjdlkfjl;jf;sfjdsl;al/ld;femflsad;fiwejdif1#incl

Problem DescriptionVerify Goldbach conjecture that any even number of sufficiently large (>=4) can be represented by a sum of two primes.InputEnter an even n. (2OutputFind A, B make n=a+bWhere A and B are two primes, and aSample Input4100Sample Output2 23 971#include 2 3#include 4 5 using namespacestd;6 7 intPrimeintm)8 9 {Ten One intI,n; A - if(m==1)return 0; - theN= (int) sqrt (Double) m); - - for(i=2; i) - + if(m%i==0)return 0; - +

+ if((m%i==0) (n%i==0)) - $ { $ -printf"%d%d\n", i,m*n/i); - the Break; - Wuyi } the - } Wu - } About $ } - -}Other code:1#include 2#include 3 intgcdintAintb)4 {5 if(a%b==0)6 returnb;7 Else8 returnGCD (b,a%b);9 }Ten intLcmintAintb) One { A returna

The title description has n integers, so that the previous number in the order to move backward m position, the last m number into the front number of m, see figure. Write a function: To achieve the above function, in the main function input n number and output adjusted n number. Enter the number of input data n n integer moved position m output moving N number sample input101 2 3 4 5 6 7 8 9 102Sample output1#include 2 #defineN 1003 4 //Move Once5MoveintA[],intN)6 {7 intI, temp = a[n-1];8

The title describes the value of sn=1!+2!+3!+4!+5!+...+n!, where n is a number. Input n output and sample input5Sample output1531#include"stdio.h"2 3 intMainintargcChar Const*argv[])4 {5 6 inti, N;7 Long Long intSn =0, factor =1;8scanf"%d", n);9 Ten for(i =1; I ) One { ASn + =factor; -Factor = factor * (i +1); - } the -printf"%lld\n", Sn); - return 0; -}1014 C Language Programming Tut