Problem 48

The series, 11+22+33+...+1010=10405071317.

Find the last ten digits of the series, 11+22+33+...+10001000.

Answer:

9110846700

Completed on Thu, 23 Jul 2015, 17:26

初步思路。能够利用元对的方式计算每个数须要乘的数

def func(a):

    for i in range(0,len(a)):

        a[i][1]*=a[i][0]

    return a

a=[[i,i] for i in range(1000,0,-1)]

result=0

while len(a)>0:

    result+=a.pop()[1]

    a=func(a)

print(result%(pow(10,10)))

更进一步，能够利用数组下标得到须要乘的数

def func(a):

    for i in range(0,len(a)):

        a[i]=a[i]*(1000-i)

    return a

a=[i for i in range(1000,0,-1)]

result=0

while len(a)>0:

    result+=a.pop()

    a=func(a)

print(result%(pow(10,10)))