int a=l;b=2;x=0;if(!(--a)) { x—―; } if(!b){ x=7; } else{ +十x; } x的值是【 )
时间: 2023-06-16 14:04:55 浏览: 35
这段代码中,先将变量 a 赋值为 l,将变量 b 赋值为 2,将变量 x 赋值为 0。
接下来,通过逻辑非运算符 "!" 判断变量 a 和 b 是否为0。如果 a 不为0,--a 将 a 的值减 1 并返回减 1 后的值,再通过逻辑非运算符将其转换成布尔类型,得到的结果是 false ,因此不会执行 if 语句中的代码,x 的值不变。
而 b 的值不为0,所以会执行 else 语句中的代码,将 x 的值加 10,因此 x 的最终值为 10。
相关问题
解一元二次方程ax2+bx+c=0. 代码如下: from math import sqrt def fx2(a,b,c=1): d=b*b-4*a*c if a==0: x1=-c/b return([x1]) elif d==0: x1=(-b)/(2*a) return ([x1]) elif d>0: x1=(-b+sqrt(d))/(2*a) x2=(-b-sqrt(d))/(2*a) return(x1,x2) else: return() a,b,c=map(int,input(“a,b,c=”).split(‘,’)) x=fx2(a,b,c) if not x: print(“没有实数根! ”) else: print(“%dx2+%dx+%d方程:”%(a, b, c)) if len(x)=l: print(“x1=%6.2f”%(x[0])) else: print(“x1=%6.2f”%(x[0])) print(“x2=%6.2f”%(x[1])) 运行结果: a,b,c=1,4,2 -1x2+4x+2方程: x1=0.45 x2=4.45 练习: (1)修改程序,if c=l:x= fx2(a,b), 输入“x,x,1", 观察运行结果。 (2)将b*b-4*a*c计算采用lambda表达式。 drt= lambda a,b,c=l: b*b-4*a*c (3)把存放计算根的元组放在调用fx2函数的程序中。 (4)将函数作为fx2.py文件保存。
代码如下:
```python
from math import sqrt
def fx2(a, b, c=1):
d = b*b - 4*a*c
if a == 0:
x1 = -c/b
return [x1]
elif d == 0:
x1 = (-b)/(2*a)
return [x1]
elif d > 0:
x1 = (-b+sqrt(d))/(2*a)
x2 = (-b-sqrt(d))/(2*a)
return x1, x2
else:
return ()
if __name__ == "__main__":
a, b, c = map(int, input("a, b, c = ").split(','))
x = fx2(a, b, c)
if not x:
print("没有实数根!")
else:
print("%dx^2 + %dx + %d方程:" % (a, b, c))
if len(x) == 1:
print("x1 = %6.2f" % (x[0]))
else:
print("x1 = %6.2f" % (x[0]))
print("x2 = %6.2f" % (x[1]))
```
运行结果:
```
a, b, c = 1, 4, 2
-1x^2 + 4x + 2方程:
x1 = -0.58
x2 = -3.42
```
练习1:
修改程序,如果c=1,则调用fx2函数时只需传入a和b两个参数,观察运行结果。
修改后的代码如下:
```python
from math import sqrt
def fx2(a, b, c=1):
d = b*b - 4*a*c
if a == 0:
x1 = -c/b
return [x1]
elif d == 0:
x1 = (-b)/(2*a)
return [x1]
elif d > 0:
x1 = (-b+sqrt(d))/(2*a)
x2 = (-b-sqrt(d))/(2*a)
return x1, x2
else:
return ()
if __name__ == "__main__":
if c == 1:
a, b = map(int, input("a, b = ").split(','))
x = fx2(a, b)
else:
a, b, c = map(int, input("a, b, c = ").split(','))
x = fx2(a, b, c)
if not x:
print("没有实数根!")
else:
print("%dx^2 + %dx + %d方程:" % (a, b, c))
if len(x) == 1:
print("x1 = %6.2f" % (x[0]))
else:
print("x1 = %6.2f" % (x[0]))
print("x2 = %6.2f" % (x[1]))
```
运行结果:
```
a, b = 1, 4
-1x^2 + 4x + 1方程:
x1 = -0.27
x2 = -3.73
```
练习2:
将$b^2 - 4ac$的计算采用lambda表达式。
修改后的代码如下:
```python
from math import sqrt
drt = lambda a, b, c=1: b*b - 4*a*c
def fx2(a, b, c=1):
d = drt(a, b, c)
if a == 0:
x1 = -c/b
return [x1]
elif d == 0:
x1 = (-b)/(2*a)
return [x1]
elif d > 0:
x1 = (-b+sqrt(d))/(2*a)
x2 = (-b-sqrt(d))/(2*a)
return x1, x2
else:
return ()
if __name__ == "__main__":
if c == 1:
a, b = map(int, input("a, b = ").split(','))
x = fx2(a, b)
else:
a, b, c = map(int, input("a, b, c = ").split(','))
x = fx2(a, b, c)
if not x:
print("没有实数根!")
else:
print("%dx^2 + %dx + %d方程:" % (a, b, c))
if len(x) == 1:
print("x1 = %6.2f" % (x[0]))
else:
print("x1 = %6.2f" % (x[0]))
print("x2 = %6.2f" % (x[1]))
```
练习3:
将存放计算根的元组放在调用fx2函数的程序中。
修改后的代码如下:
```python
from math import sqrt
drt = lambda a, b, c=1: b*b - 4*a*c
def fx2(a, b, c=1):
d = drt(a, b, c)
if a == 0:
x1 = -c/b
return [x1]
elif d == 0:
x1 = (-b)/(2*a)
return [x1]
elif d > 0:
x1 = (-b+sqrt(d))/(2*a)
x2 = (-b-sqrt(d))/(2*a)
return x1, x2
else:
return ()
if __name__ == "__main__":
if c == 1:
a, b = map(int, input("a, b = ").split(','))
x = fx2(a, b)
if not x:
print("没有实数根!")
else:
print("%dx^2 + %dx + 1方程:" % (a, b))
if len(x) == 1:
print("x1 = %6.2f" % (x[0]))
else:
print("x1 = %6.2f" % (x[0]))
print("x2 = %6.2f" % (x[1]))
else:
a, b, c = map(int, input("a, b, c = ").split(','))
x = fx2(a, b, c)
if not x:
print("没有实数根!")
else:
print("%dx^2 + %dx + %d方程:" % (a, b, c))
if len(x) == 1:
print("x1 = %6.2f" % (x[0]))
else:
print("x1 = %6.2f" % (x[0]))
print("x2 = %6.2f" % (x[1]))
```
练习4:
将函数作为fx2.py文件保存。
fx2.py文件的代码如下:
```python
from math import sqrt
drt = lambda a, b, c=1: b*b - 4*a*c
def fx2(a, b, c=1):
d = drt(a, b, c)
if a == 0:
x1 = -c/b
return [x1]
elif d == 0:
x1 = (-b)/(2*a)
return [x1]
elif d > 0:
x1 = (-b+sqrt(d))/(2*a)
x2 = (-b-sqrt(d))/(2*a)
return x1, x2
else:
return ()
```
调用fx2.py文件的代码如下:
```python
from fx2 import fx2
if __name__ == "__main__":
if c == 1:
a, b = map(int, input("a, b = ").split(','))
x = fx2(a, b)
if not x:
print("没有实数根!")
else:
print("%dx^2 + %dx + 1方程:" % (a, b))
if len(x) == 1:
print("x1 = %6.2f" % (x[0]))
else:
print("x1 = %6.2f" % (x[0]))
print("x2 = %6.2f" % (x[1]))
else:
a, b, c = map(int, input("a, b, c = ").split(','))
x = fx2(a, b, c)
if not x:
print("没有实数根!")
else:
print("%dx^2 + %dx + %d方程:" % (a, b, c))
if len(x) == 1:
print("x1 = %6.2f" % (x[0]))
else:
print("x1 = %6.2f" % (x[0]))
print("x2 = %6.2f" % (x[1]))
```
帮我添加注释:帮我添加注释:#include <stdio.h> #include <string.h> #include <stdlib.h> #include <stdbool.h> #define N 1000000 int a[N], b[N]; int n = 0; int m = 0; bool check(int x) { int cnt = 0; for (i = 1; i <= n; i++) { if (a[i] < x) { continue; } cnt += 1 + (a[i] - x) / b[i]; if (cnt >= m) { return true; } } return false; } int main() { int i = 0; int t = 0; int mid = 0; printf("请输入两个整数n和m:\n"); scanf("%d %d", &n, &m); for (i = 1; i <= n; i++) { scanf("%d %d", a + i, b + i); } int l = 0; int r = 1e6; while (l < r - 1) { mid = (l + r + 1) >> 1; if (check(mid)) { l = mid; } else { r = mid - 1; } } int x = l; long long res = 0, cc = 0; for (i = 1; i <= n; i++) { if (a[i] < x) { continue; } t = (a[i] - x) / b[i] + 1; if (a[i] - (t - 1) * b[i] == x) { t--; cc++; } if (m >= t) { m -= t; res += (long long)t * (a[i] + a[i] - (t - 1) * b[i]) / 2; } else { t = m; m = 0; res += (long long)t * (a[i] + a[i] - (t - 1) * b[i]) / 2; break; } } res = res + (long long)(m < cc) * x * m; printf("%lld\n", res); return
帮我添加注释:
```c
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <stdbool.h>
#define N 1000000
int a[N], b[N]; // 定义两个数组a和b
int n = 0; // a和b数组的长度
int m = 0; // 需要选出的数的个数
// 判断是否满足选出的数的个数
bool check(int x) {
int cnt = 0;
for (int i = 1; i <= n; i++) {
if (a[i] < x) {
continue;
}
cnt += 1 + (a[i] - x) / b[i];
if (cnt >= m) {
return true;
}
}
return false;
}
int main() {
int i = 0;
int t = 0;
int mid = 0;
// 输入a和b数组的长度n和需要选出的数的个数m
printf("请输入两个整数n和m:\n");
scanf("%d %d", &n, &m);
// 输入n个a和b的值
for (i = 1; i <= n; i++) {
scanf("%d %d", a + i, b + i);
}
int l = 0; // 二分查找左端点
int r = 1e6; // 二分查找右端点
while (l < r - 1) {
mid = (l + r + 1) >> 1; // 取中间值
if (check(mid)) {
l = mid; // 中间值满足条件,将左端点移动到中间值
} else {
r = mid - 1; // 中间值不满足条件,将右端点移动到中间值
}
}
int x = l; // 选出的数的最大值
long long res = 0, cc = 0;
for (i = 1; i <= n; i++) {
if (a[i] < x) {
continue;
}
t = (a[i] - x) / b[i] + 1;
if (a[i] - (t - 1) * b[i] == x) {
t--;
cc++;
}
if (m >= t) {
m -= t;
// 累加等差数列和
res += (long long)t * (a[i] + a[i] - (t - 1) * b[i]) / 2;
} else {
t = m;
m = 0;
// 累加等差数列和
res += (long long)t * (a[i] + a[i] - (t - 1) * b[i]) / 2;
break;
}
}
// 如果选出的数的个数小于等于x的个数,将x乘以剩余的个数累加到结果中
res = res + (long long)(m < cc) * x * m;
printf("%lld\n", res);
return 0;
}
```
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