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#include <bits/stdc++.h>
using namespace std;
int main() {
int n = 1, ans = 0;
for (int i = 1; i <= 2019; i++) n = n * 2019 % 100; //计算出n'
// 1~11的n次方
for (int i = 1; i <= 11; i++) {
int x = 1;
for (int j = 1; j <= n; j++) x = x * i % 101; //降幂后可以正常按要求计算
//收集答案
ans += x;
}
//最终也要模一下101
printf("%d\n", ans % 101);
return 0;
}