就点个赞吧~~求了
2025-07-20 10:56:37
发布于:浙江
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这道题我写了好就,点个赞吧
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<bits/stdc++.h>
using namespace std;
int n,m1,m2,ans = 150000;
int i,j;
int s[10001],f[10001];
bool p[30001];
int prime[501][3],size = 0;
int main(void) {
memset(p,1,sizeof(p));
memset(f,0,sizeof(f));
scanf("%d%d%d",&n,&m1,&m2);
if (m1 == 1) {
printf("0");
return 0;
}
for (i = 1; i <= n; i++)
scanf("%d",&s[i]);
int xx = floor(sqrt(m1));
for (i = 2; i <= xx; i++) {
if (p[i]) {
if (m1 % i == 0) {
prime[++size][1] = i;
prime[size][2] = 1;
}
}
int tim = 2;
while (tim * i <= m1) {
p[tim * i] = 0;
tim++;
}
}
for (i = 1; i <= size; i++) {
int num = prime[i][1];
while (m1 % (num * prime[i][1]) == 0) {
num *= prime[i][1];
prime[i][2]++;
}
prime[i][2] *= m2;
}
if (size == 0) {
prime[++size][1] = m1;
prime[size][2] = m2;
}
for (i = 1; i <= n; i++) {
for (j = 1; j <= size; j++) {
if (s[i] % prime[j][1] != 0) {
f[i] = 150000;
break;
}
int tim = 1;
long long num = prime[j][1];
while (s[i] % (num * prime[j][1]) == 0) {
num *= prime[j][1];
tim++;
}
int an = (prime[j][2]-1) / tim + 1;
if (an > f[i]) f[i] = an;
}
}
for (i = 1; i <= n; i++)
if (ans > f[i]) ans=f[i];
if (ans == 150000) printf("-1");
else printf("%d",ans);
return 0;
}
全部评论 3
AC了
2025-08-31 来自 广东
1是个人物
2025-08-23 来自 浙江
1#include <iostream>
#include <vector>
#include <unordered_map>
#include <climits>
using namespace std;// 质因数分解函数
unordered_map<int, int> factorize(int n) {
unordered_map<int, int> factors;
while (n % 2 == 0) {
factors[2];
n /= 2;
}
for (int i = 3; i * i <= n; i += 2) {
while (n % i == 0) {
factors[i];
n /= i;
}
}
if (n > 1) {
factors[n]++;
}
return factors;
}int main() {
int N;
cin >> N;int m1, m2; cin >> m1 >> m2; // 计算M = m1^m2的质因数分解 auto m1_factors = factorize(m1); unordered_map<int, int> M_factors; for (auto& [p, exp] : m1_factors) { M_factors[p] = exp * m2; } // 如果m1为1,那么M=1,任何细胞都能立即均分 if (m1 == 1) { cout << 0 << endl; return 0; } vector<int> S(N); for (int i = 0; i < N; i++) { cin >> S[i]; } int min_time = INT_MAX; for (int s : S) { // 分解S_i的质因数 auto s_factors = factorize(s); // 检查S_i是否包含M的所有质因数 bool valid = true; for (auto& [p, exp] : M_factors) { if (s_factors.find(p) == s_factors.end()) { valid = false; break; } } if (!valid) { continue; // 这种细胞无法满足要求 } // 计算最少需要的分裂时间 int max_t = 0; for (auto& [p, exp_M] : M_factors) { int exp_S = s_factors[p]; // 计算需要的分裂次数(向上取整) int t = (exp_M + exp_S - 1) / exp_S; max_t = max(max_t, t); } if (max_t < min_time) { min_time = max_t; } } if (min_time == INT_MAX) { cout << -1 << endl; // 没有找到合适的细胞 } else { cout << min_time << endl; } return 0;}
2025-09-24 来自 浙江
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