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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#658024#9248. An Easy Math ProblemVegetog#AC ✓176ms3936kbC++204.1kb2024-10-19 16:00:202024-10-31 22:08:50

Judging History

你现在查看的是测评时间为 2024-10-31 22:08:50 的历史记录

  • [2024-10-31 22:41:58]
  • 自动重测本题所有获得100分的提交记录
  • 测评结果:AC
  • 用时:172ms
  • 内存:3892kb
  • [2024-10-31 22:36:43]
  • hack成功,自动添加数据
  • (/hack/1098)
  • [2024-10-31 22:19:06]
  • 自动重测本题所有获得100分的提交记录
  • 测评结果:100
  • 用时:176ms
  • 内存:4012kb
  • [2024-10-31 22:13:58]
  • hack成功,自动添加数据
  • (/hack/1096)
  • [2024-10-31 22:08:50]
  • 自动重测本题所有获得100分的提交记录
  • 测评结果:100
  • 用时:176ms
  • 内存:3936kb
  • [2024-10-31 22:00:43]
  • hack成功,自动添加数据
  • (/hack/1095)
  • [2024-10-19 16:00:20]
  • 评测
  • 测评结果:100
  • 用时:175ms
  • 内存:3972kb
  • [2024-10-19 16:00:20]
  • 提交

answer

#include<bits/stdc++.h>
#define int long long
using namespace std;
using ll =long long;
const int N=2e5+6;

namespace prime_fac {
const int S = 8; // 随机算法判定次数,8~10 就够了

// 龟速乘
    long long mult_mod(long long a, long long b, long long c) {
        a %= c, b %= c;
        long long ret = 0;
        long long tmp = a;
        while (b) {
            if (b & 1) {
                ret += tmp;
                if (ret > c) ret -= c;
            }
            tmp <<= 1;
            if (tmp > c) tmp -= c;
            b >>= 1;
        }
        return ret;
    }

    // 快速幂
    long long qow_mod(long long a, long long n, long long _mod) {
        long long ret = 1;
        long long temp = a % _mod;
        while (n) {
            if (n & 1) ret = mult_mod(ret, temp, _mod);
            temp = mult_mod(temp, temp, _mod);
            n >>= 1;
        }
        return ret;
    }

    // 是合数返回true,不一定是合数返回false
    bool check(long long a, long long n, long long x, long long t) {
        long long ret = qow_mod(a, x, n);
        long long last = ret;
        for (int i = 1; i <= t; i++) {
            ret = mult_mod(ret, ret, n);
            if (ret == 1 && last != 1 && last != n - 1) return true;
            last = ret;
        }
        if (ret != 1) return true;
        return false;
    }

    // 是素数返回true,不是返回false
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    bool Miller_Rabin(long long n) {
        if (n < 2) return false;
        if (n == 2) return true;
        if ((n & 1) == 0) return false;
        long long x = n - 1;
        long long t = 0;
        while ((x & 1) == 0) { x >>= 1; t++; }

        for (int i = 0; i < S; i++) {
            long long a = rng() % (n - 1) + 1;
            if (check(a, n, x, t))
                return false;
        }

        return true;
    }

    long long factor[100];// 存质因数
    int tol; // 质因数的个数,0~tol-1

    long long gcd(long long a, long long b) {
        long long t;
        while (b) {
            t = a;
            a = b;
            b = t % b;
        }
        if (a >= 0) return a;
        return -a;
    }

    long long pollard_rho(long long x, long long c) {
        long long i = 1, k = 2;
        long long x0 = rng() % (x - 1) + 1;
        long long y = x0;
        while (1) {
            i++;
            x0 = (mult_mod(x0, x0, x) + c) % x;
            long long d = gcd(y - x0, x);
            if (d != 1 && d != x) return d;
            if (y == x0) return x;
            if (i == k) { y = x0; k += k; }
        }
    }
    // 对n质因数分解,存入factor,k一般设置为107左右
    void findfac(long long n, int k) {
        if (n == 1) return;
        if (Miller_Rabin(n)) {
            factor[tol++] = n;
            return;
        }
        long long p = n;
        int c = k;
        while (p >= n) p = pollard_rho(p, c--);
        findfac(p, k);
        findfac(n / p, k);
    }
    vector<int> fac(long long n) {
        tol = 0;
        vector<int>ret;
        findfac(n, 107);
        for (int i = 0; i < tol; i++)ret.push_back(factor[i]);
        return ret;
    }
}
int b[20];
void solve()
{
    ll n,m=0,x;
    cin>>n;
    vector<int> fac = prime_fac::fac(n);
    vector<ll> a;
    unordered_map<ll, ll> mp;
    for(auto i : fac) {
        if(mp[i] == 0) 
            a.push_back(i);
        mp[i] ++ ;
    }
    ll ans = 1, res = 0;   
    int cp=pow(2,(a.size()));
    for(int i=1;i<cp;i++){
        int er=i,w=0;
        while(er){
            if(er%2){
                b[++w]=1;
            }else b[++w]=0;
            er/=2;
        }
        res=1;
        for(int j=1;j<=w;j++){
            if(b[j]==0)continue;
            res*=mp[a[j-1]]*2;
        }
        res/=2;
        ans+=res;

    }

    cout << ans << "\n";
}

signed main(){
    ios::sync_with_stdio(0);
    cin.tie(nullptr),cout.tie(nullptr);

    int T=1;
    cin>>T;
    while(T--){
        solve();
    }
}

这程序好像有点Bug,我给组数据试试?

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 3784kb

input:

10
1
2
3
4
5
6
7
8
9
10

output:

1
2
2
3
2
5
2
4
3
5

result:

ok 10 lines

Test #2:

score: 0
Accepted
time: 176ms
memory: 3936kb

input:

2000
6469693230
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6469693230
6469693230
646969323...

output:

29525
29525
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29525
29525
29525
29525
29525
29525
29525
29525
29525
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29525
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29525
...

result:

ok 2000 lines

Test #3:

score: 0
Accepted
time: 84ms
memory: 3932kb

input:

2000
1763047095
79735483
1016286871
2864801397
2327774116
2668010360
3469893354
3634459021
1613699068
781737219
574741575
2763134701
1458502604
1822260248
2281150332
2924219311
2493931196
3735904708
158802001
2006921221
729928782
1974841034
727412600
2873393292
1291087179
2741607663
1893408215
29827...

output:

14
5
2
5
23
95
68
14
8
68
203
14
23
32
38
41
8
8
14
2
608
41
158
338
23
41
14
5
14
41
14
203
41
14
17
446
5
53
59
878
2
14
365
203
14
203
2
122
32
95
41
41
5
23
14
41
5
5
14
122
23
203
608
23
41
122
2
14
95
2
68
41
203
14
230
41
68
23
50
14
32
14
8
5
5
5
68
68
122
293
473
5
41
41
14
2
14
14
5
2
122
...

result:

ok 2000 lines

Extra Test:

score: 0
Extra Test Passed