求解1-3+5-7+9-11的值

n=int(input('请输入整数n：'))
i=0
sum=0
a=0
while i<=n:
    if i>=1 and i%4==1:
        sum+=i
    elif i%2!=0 and i!=1:
        a=a+i
    i+=1
print(sum-a)

你题目的解答代码如下：（如有帮助，望采纳！谢谢! 点击我这个回答右上方的【采纳】按钮）

s = 0
t = ""
n=int(input('请输入整数n：'))
for i in range(1, n+1, 2):
    if i%4==1:
        s += i
        t += "+"+str(i)
    else:
        s -= i
        t += "-"+str(i)
print("{}={}".format(t[1:],s))

n = int(input("请输入整数n"))
f = 1
s = 0

for i in range(1,n+1,2) :
    s += f * i
    f = -f

print(s)

def js(n):
    s=0
    s_str=''
    jia=True
    for i in range(1,n+1):
        if i%2==0:
            continue
        if jia:
            s+=i
            if s_str=='':
                s_str+=str(i)
            else:
                s_str+='+'+str(i)
            jia=False
        else:
            s-=i
            s_str+='-'+str(i)
            jia=True
    return s_str+'='+str(s)
n=input('请输入整数n：')
print(js(int(n))) # 1-3+5-7+9-11=-6

n = int(input("请输入整数n："))
sum = 0
sign = 1
 
for i in range(1, n+1, 2) :
    sum += sign * i
    sign = -sign
 
print(sum)