Problem 48 - Find the last ten digits of 1^1 + 2^2 + ... + 1000^1000.

링크

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

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

python

str(sum(x**x for x in range(1, 1001)))[-10:]

c++

#include <cstdlib> #include <cstdio> int main() { long long mod = 10000000000LL; long long ans = 0; for (int i = 1; i <= 1000; i++) { long long p = 1; for (int j = 0; j < i; j++) { p = (p * i) % mod; } ans = (ans + p) % mod; } printf("%lld\n", ans); return 0; }