結果

問題 No.375 立方体のN等分 (1)
ユーザー FF256grhy
提出日時 2016-06-07 11:14:53
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 5 ms / 5,000 ms
コード長 2,215 bytes
コンパイル時間 265 ms
コンパイル使用メモリ 24,832 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-10-11 21:05:19
合計ジャッジ時間 1,289 ms
ジャッジサーバーID
(参考情報)
judge1 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 32
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:100:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  100 |         scanf("%lld", &n);
      |         ~~~~~^~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <stdio.h>
typedef long long int LL;
LL n, factor[12], expo[12], length;
void factorize() {
LL m = n;
for(LL i = 2; i * i <= m; i++) {
if(m % i == 0) {
factor[length] = i;
while(m % i == 0) {
m /= i;
expo[length]++;
}
length++;
}
}
if(m != 1) {
factor[length] = m;
expo[length]++;
length++;
}
return;
}
int size() {
int ans = 1;
for(int i = 0; i < length; i++) { ans *= expo[i] + 1; }
return ans;
}
LL val(int v) {
LL ans = 1;
for(int i = 0; i < length; i++) {
for(int j = 0; j < v % (expo[i] + 1); j++) {
ans *= factor[i];
} v /= (expo[i] + 1);
}
return ans;
}
void sort(LL* p, LL* q, int l, int r) {
if(r - l < 2) { return; }
int m = (l + r) / 2;
sort(p, q, l, m);
sort(p, q, m, r);
int ll = l, rr = m, c = l;
while(ll < m || rr < r) {
bool flag;
if(ll == m) { flag = false; }
else if(rr == r) { flag = true; }
else { flag = (p[ll] <= p[rr]); }
if(flag) { q[c++] = p[ll++]; }
else { q[c++] = p[rr++]; }
}
for(c = l; c < r; c++) { p[c] = q[c]; }
return;
}
LL sqrt(LL x) {
LL min = 0, max = 10000000;
while(max - min > 1) {
LL mid = (max + min) / 2;
if(mid * mid <= x) { min = mid; } else { max = mid; }
}
return min;
}
LL cbrt(LL x) {
LL min = 0, max = 100000;
while(max - min > 1) {
LL mid = (max + min) / 2;
if(mid * mid * mid <= x) { min = mid; } else { max = mid; }
}
return min;
}
int find(LL* p, int s, LL x) {
int min = 0, max = s;
while(max - min > 1) {
int mid = (max + min) / 2;
if(p[mid] <= x) { min = mid; } else { max = mid; }
}
return min;
}
int main() {
scanf("%lld", &n);
factorize();
int s = size();
LL* p = new LL[s];
LL* q = new LL[s];
for(int i = 0; i < s; i++) {
p[i] = val(i);
}
sort(p, q, 0, s);
delete[] q;
int cr = find(p, s, cbrt(n));
LL min = p[cr] + n / p[cr] + 1;
for(int i = cr; 0 <= i && p[i] + 2 * sqrt(n / p[i]) < min; i--) {
for(int j = find(p, s, sqrt(n / p[i])); 0 <= j && p[i] + p[j] + n / p[i] / p[j] < min; j--) {
if(n / p[i] % p[j] == 0) {
LL k = p[i] + p[j] + n / p[i] / p[j];
if(k < min) { min = k; }
}
}
}
printf("%lld %lld\n", min - 3, n - 1);
return 0;
}
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