Convert pseudocode to c++ code PO: T1 = Time(); for (i = 0; i < 1000; i++) send(x[i],P1); T2 = Time(): TPO = T2-T1: Recv(TP1, P1); Print(TPO): Print(TP1); P1: T1 = Time(): for (i = 0; i < 1000; i++) recv(x[i].PO): T2 = Time(): TP1 = T2- T1: send(TP1, PO); 7-0
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![Convert pseudocode to c++ code
PO:
T1 = Time();
for (i = 0; i < 1000; i++)
send(x[i],P1);
T2 = Time();
TPO = T2-T1;
Recv(TP1, P1);
Print(TPO);
Print(TP1);
P1:
T1 = Time();
for (i = 0; i < 1000; i++)
recv(x[i],PO);
T2 Time();
=
TP1 = T2-T1;
send(TP1, P0);
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- int functionC (int n) { int i, j, sumC = 0; for (i=n; i > 0; i=i-5) for (j=1; j 0) { if (functionC(n) % 2 == 0) { for (i=m; i > 0; i=i/3) sumE++; } else (10) Asymptotic runtime of functionE { for (i=m; i > 0; i=i-3) sumE--; } n--; } return sumE;Convert pseudocode to c++ code PO: T1 = Time(); for (i = 0; i < 1000; i++) send(x[i],P1); T2 = Time(); TPO = T2-T1; Recv(TP1, P1); Print(TPO); Print(TP1); P1: T1 = Time(); for (i = 0; i < 1000; i++) recv(x[i],PO); T2 = Time(); TP1 = T2-T1; send(TP1, P0);FIX ME: In C language Update this function to display the string of 16 bits using the following pattern: 11111 111111 11111 (5) (6) (5) */ void printbits(unsigned short wrd) { int testbit(unsigned short wrd, int bit_to_test); int i; for (i = 15; i >= 0; i--) { printf("%1d", testbit(wrd, i)); if (!(i % 4)) printf(" "); } printf("\n"); }
- For the C code int dw_loop(long x) { long y = x* x; long 'p - &x; long n - 2*x do { x += y; (*p)++; n--; } while (n > 0); return x; }Please convert to C language #include <bits/stdc++.h>using namespace std; //folding method int main(){ string s; cout << "enter number\n"; cin >> s; //folding and summing int sum = 0; for (int i = 0; i < s.length(); i += 2) { if (i + 1 < s.length()) sum += stoi(s.substr(i, 2)); else //when only one digit is left for folding sum += stoi(s.substr(i, 1)); } cout << s << "->" << sum % 10; return 0;} Output:write in c language Note:the sample input is: And the sample output is ; 3 6 9 3 10 15 20 5 Not 3 6 9 3 both inputs need to be in the output Description 請寫一個程式找到三個數的最大公因數。 Please write a program to find the greatestcommon divisor of three number. Input 輸入會包含多筆測試資料，每筆測試資料一行，每行中有三個正整數，整數間會用一個空白隔開。 Input consist several test case, each test case a line. For every test case there are three integer in it and separate by a space. Output 對每一筆測試資料輸出一行結果。 For each test case output the result in one line. Sample Input 1. Sample Output 1 3 6 9 3 10 15 20 5
- Computer Science Implement the Histogram Equalization according to the explanation in Wikipedia in C# Starting code: public static bool ConvertToGray(Bitmap b) { return true; }Write in C language Description 給你十個數字，請從小到大輸出這十個數字。 Giving 10 number, output them in ascending order. Input 輸入會包含十個整數，數字間會用一個空白隔開。 Input consist 10 integer, each of them separate by a single space. Output 從小到大輸出十個數 output the 10 number inascending order. Sample Input 1 9 8 7 6 5 4 3 2 1 0 Sample Output 1 0 1 2 3 4 5 6 7 8 9Write in C Language Description F(x) = x + 1 G(x, y) = x + y Please write a program to calculate the value of composite of F and G Input Input will be the composite of F and G. The parameter of F and G are integer only. Output An integer represent the value of function. Sample Input 1 F(G(1,F(3))) Sample Output 1 6
- In C code This is what I have so far #include <stdio.h> int main () { unsigned int x = 0x76543210; char *c = (char*) &x; printf ("*c is: 0x%x\n", *c); if (*c == 0x10) { printf ("Architecture is little endian. \n"); } else { printf ("Architecture is big endian. \n"); } return 0; }Write a program in C++ to find the Greatest Common Divisor (GCD) of two numbers iteratively.Sample Output:Input the first number: 12Input the second number: 8The Greatest Common Divisor is: 4JAVA and Netbeans Using Looping Statement, Create a simple Multiplication Table for user input number (same as below):