Project Euler #254: Sums of Digit Factorials hacerrank code example
Example: Project Euler #254: Sums of Digit Factorials
using namespace std;
using namespace fmt;
using namespace NTL;
namespace PE254 {
const int MAX_LENGTH = 18, MAX_SUM = 150, MOD = 362880;
unsigned int f[MAX_LENGTH + 1][MAX_SUM + 1][MOD/32 + 100];
long power[MAX_LENGTH + 2];
long backtrace(int len, int sum, int m) {
long ret = 0;
for (; len; ) {
for (int i = 0; i < 10 && i <= sum; ++i) {
int pos = (m + (MOD - i) * power[len - 1]) % MOD;
if (getbit(f[len - 1][sum - i], pos)) {
ret = ret * 10 + i;
len -= 1;
sum -= i;
m = pos;
}
}
}
return ret;
}
struct node {
long x[10];
node() {
for (int i = 0; i < 10; ++i)
x[i] = 0;
}
long sum() {
long ret = 0;
for (int i = 0; i < 10; ++i)
ret += x[i];
return ret;
}
};
bool operator < (node a, node b) {
long sa = a.sum(), sb = b.sum();
if (sa != sb) return sa < sb;
for (int i = 0; i < 10; ++i)
if (a.x[i] != b.x[i])
return a.x[i] > b.x[i];
return false;
}
long calc(long s) {
node ret; long y = 0;
ret.x[9] = 1e18;
for (int m = 0; m < MOD; ++m) {
for (int i = 1; i <= MAX_LENGTH; ++i) {
if (getbit(f[i][s], m)) {
//if (m == 5762)
long Y = backtrace(i, s, m), Y_old = Y;
node t;
t.x[9] += Y/MOD, Y %= MOD;
for (int fac = MOD, k = 8; k > 0; --k) {
fac/= k + 1;
t.x[k] += Y/fac;
Y %= fac;
}
//if (t.x[1] > 1) {
// t.x[0] = t.x[1] - 1;
// t.x[1] = 1;
//}
if (t < ret) {
ret = t;
y = Y_old;
}
break;
}
}
}
long sum = 0;
for (int i = 0; i <= 9; ++i)
sum += i * ret.x[i];
print("Y = {}\n", y);
return sum;
}
void main() {
setbit(f[0][0], 0);
power[0] = 1;
for (int i = 0; i < 18; ++i)
power[i + 1] = power[i] * 10 % MOD;
for (int i = 0; i < MAX_LENGTH; ++i) {
for (int s = 0; s <= MAX_SUM; ++s) {
for (int m = 0; m < MOD; ++m) {
if (getbit(f[i][s], m)) {
for (int t = 0; t <= 9; ++t) {
if (s + t <= MAX_SUM) {
int pos = (m * 10 + t) % MOD;
setbit(f[i + 1][s + t], pos);
}
}
}
}
}
}
ZZ ans{0};
for (int i = 1; i <= 150; ++i) {
long cur = calc(i);
print("{}: {}\n", i, cur);
ans += cur;
}
print("ans = {}\n", ans);
}
}
int main() {
PE254::main();
return 0;
}