725

Gn_6

package main
import "fmt"

func main(){
var time int =0
fmt.Scanf("%d", &time)
fmt.Printf("%d:%d:%d", time/3600, time%3600/60, time%60)
}

#### Go 代码

package main
import "fmt"

func main(){
money := 0
var hundred,fifty,twenty,ten,five,two,one int = 0,0,0,0,0,0,0
fmt.Scanf("%d", &money)
hundred = money/100
fifty = money%100/50
twenty = (money-(money/50)*50)/20
ten = (money- (money/50)*50 - ((money-(money/50)*50)/20)*20  )/10
five = money%10/5
two = (money-(money/5)*5)/2
one = money- (money/5)*5 - ((money-(money/5)*5)/2)*2
fmt.Printf("%d\n"        +
"%d nota(s) de R$100,00\n"+ "%d nota(s) de R$ 50,00\n" +
"%d nota(s) de R$20,00\n" + "%d nota(s) de R$ 10,00\n" +
"%d nota(s) de R$5,00\n" + "%d nota(s) de R$ 2,00\n"  +
"%d nota(s) de R$1,00\n", + money, hundred, fifty, twenty, ten, five, two, one) } ### 题目描述 请编写一个程序，可以读取一名员工的员工编号，本月工作总时长（小时）以及时薪，并输出他的工资条，工资条中包括员工编号和员工月收入。 ### 输入格式 输入包含两个整数和一个浮点数，分别代表员工编号，工作时长以及时薪。 每个数占一行。 ### 输出格式 输出共两行，第一行格式为 NUMBER = X，其中 X 为员工编号。 第二行格式为 SALARY = U$ Y，其中 Y 为该员工月收入，保留两位小数。

$1 \leq 员工编号 \leq 100$,
$1 \leq 总工作时长 \leq 200$,
$1\leq 时薪 \leq 50$

25
100
5.50

### 输出样例：

NUMBER = 25
SALARY = U$550.00 #### GO 代码 package main import "fmt" func main(){ var num,time int = 0,0 var money float64 = 0.0 fmt.Scanf("%d\n%d\n%f", &num, &time, &money) fmt.Printf("NUMBER = %d\nSALARY = U$ %.2f", num, money*float64(time))
}

9个月前
#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
int n;
cin >> n;
cout << n << endl;
printf("%d nota(s) de R$100,00\n", n/100); n %= 100; printf("%d nota(s) de R$ 50,00\n", n/50); n %= 50;
printf("%d nota(s) de R$20,00\n", n/20); n %= 20; printf("%d nota(s) de R$ 10,00\n", n/10); n %= 10;
printf("%d nota(s) de R$5,00\n", n/5); n %= 5; printf("%d nota(s) de R$ 2,00\n", n/2); n %= 2;
printf("%d nota(s) de R$1,00\n", n/1); n %= 1; return 0; } 活动打卡代码 AcWing 616. 两点间的距离 王不留行 9个月前 #include <iostream> #include <cstdio> #include <cmath> using namespace std; int main(int argc, char* argv[]) { double p1, p2, p3, p4; cin >> p1 >> p2 >> p3 >> p4; double x1 = p3 - p1, x2 = p4 - p2; printf("%.4f", sqrt(x1 * x1 + x2 * x2)); return 0; } 活动打卡代码 AcWing 615. 油耗 王不留行 9个月前 #include <iostream> #include <cstdio> using namespace std; int main(int argc, char* argv[]) { int road; double oil; cin >> road >> oil; printf("%.3f km/l", 1.0*road/oil); return 0; } 活动打卡代码 AcWing 609. 工资 王不留行 9个月前 #include <iostream> #include <cstdio> using namespace std; int main(int argc, char* argv[]) { int id, time; double pay; cin >> id >> time >> pay; printf("NUMBER = %d\nSALARY = U$ %.2f", id, pay*time);
return 0;
}

9个月前
#include <iostream>
#include <cstdio>

using namespace std;

int main(int argc, char* argv[])
{
double a, b;
cin >> a >> b;
//cout << "A=" << PI*r*r;
printf("MEDIA = %.5f", (a*3.5+b*7.5)/11.0);
return 0;
}

9个月前
#include <iostream>
#include <cstdio>

using namespace std;

int main(int argc, char* argv[])
{
const double PI = 3.14159;
double r;
cin >> r;
//cout << "A=" << PI*r*r;
printf("A=%.4f", PI*r*r);
return 0;
}

9个月前
#include <iostream>

using namespace std;

int main(int argc, char* argv[])
{
int a, b, c, d;
cin >> a >> b >> c >> d;
cout << "DIFERENCA = " << a*b - c*d;
return 0;
}