Problem 48 | Project Euler |

[sx349]

https://pe-cn.github.io/48/

Problem 48 Self powers The series, 11 + 22 + 33 + … + 1010 = 10405071317. Find the last ten digits of the series, 11 + 22 + 33 + … + 10001000. 自幂 十项的自幂级数求和为 11 + 22 + 33 + … + 1010 = 10405071317。 求如

[hyf2018gdfz]

即让答案对 10^10 取模

#include<bits/stdc++.h>
#define mod 10000000000
int main(){
    long long ans = 0;
    for(int i = 1;i<=1000;++i){
        long long tmp = 1;
        for(int j = 1;j<=i;++j) tmp = tmp*i%mod;
        ans = (ans+tmp)%mod;
    }
    return printf("%lld",ans),0;
}

[shangkelingxiang]

ans=0
p=10**10
for i in range(1,1001):
    ans=(ans+pow(i,i,p))%p
print(ans)

[p-gp]

#include <stdio.h>

long long int selfpow(int x);

int main()
{
    long long int sum = 0;
    for (int i = 1; i <= 1000; ++i) {
        sum += selfpow(i);
        sum %= 10000000000;
    }
    printf("%lld\n", sum);

    return 0;
}

long long int selfpow(int x) {
    int i;
    long long int y = (long long)x;
    for (i = 1; i < x; ++i) {
        y *= (long long)x;
        y %= 10000000000;
    }

    return y;
}

[SunMoonTrain]

FromDigits@Take[IntegerDigits@Sum[n^n, {n, 1000}], -10]