(-1)+(2)+(-3)+(+4)+......+(-2003)+(+2004)的答案是什么

1002
一对数是1，总共可看做1002对，所以是1002

sum = 0
for integer in range( 1, 2005 ):
sum1 = pow( ( -1 ), ( integer ) ) * integer
sum += sum1
print "%d" % sum

sum = 1002

先算2+...+2004 =(2+2004)*502/2=
再算1+...+2003=(1+2003)*502/2=
把两个结果相减就行了

-(1+3+..+2003)+(2+4+..+2004)=-1002x2004/2+1002x2005/2=501

(-1+2)=1,-3+4=1,以此类推共有1002个1，则为1002.