[交流] Project Euler 48 欧拉工程 48 题 已有5人参与

21楼: Originally posted by wangww2011 at 2011-09-08 21:55:35:
楼上的给出一个了，不用sin,cos也一样

a(2*n-1)=21+n
a(2*n)=28-5/2*n+(1/4)*(-1)^n+1/4
====>
[code]
a(n)=(1+(-1)^(n+1))/2*(21+(n+1)/2)+(1+(-1)^n)/2*(28-5/4*n-1/4*(-1)^(n/2)+1/4)
[/code ...

print sum([i**i for i in xrange(1,1001)])%10000000000

print sum([(i**i)%10000000000 for i in xrange(1,1001)])%10000000000

#include

int main()
{
    unsigned int i, j;
    unsigned long long sum=0, tmp;
    const unsigned long long k=10000000000ULL;
    for(i=1; i<=1000; i++){
        tmp=i;
        for(j=1; j             tmp *= i;
            tmp %= k;
        }
        sum += tmp;
    }

    printf("%010I64u\n", sum%k);

    return 0;
}