/* next function is adapted from:
Inline C++ integer exponentiation routines
Version 1.01
Copyright (C) 1999-2004 John C. Bowman <*****@*****.**>
*/
static inline Int
ipow(Int x, Int p)
{
Int r;
if (p == 0) return ((CELL)1);
if (x == 0 && p > 0) return 0L;
if(p < 0)
return (-p % 2) ? x : ((CELL)1);
r = ((CELL)1);
for(;;) {
if(p & 1) {
if (mul_overflow((r*x), r, x)) {
return 0;
}
r *= x;
}
if((p >>= 1) == 0) return r;
if (mul_overflow((x*x), x, x)) {
return 0;
}
x *= x;
}
}
__int64 solve_119()
{
std::list<ui64_t> xlist;
for(int base = 2; base < 500; ++base)
{
// 乘方的方式构造数,power必然是完全N次方数,需要检验的是它的DigitSum是否和它N次方的基数相等.
for(ui64_t power = base * base; !mul_overflow(power, base); power *= base)
if(digit_sum(power) == base) // 按顺序插入.
xlist.insert(std::lower_bound(xlist.begin(), xlist.end(), power), power);
}
int i = 0;
std::list<ui64_t>::const_iterator ite = xlist.begin();
for(; ite != xlist.end() && ++i < 30; ++ite)
;
return ite != xlist.end() ? *ite : 0;
}
