Solution Uva: 10338 - Mischievous Children

 

Problem	Verdict	Lang	Time	Best	Rank	Submit Time
| discuss10338 - Mischievous Children	Accepted	Python	1.020	0.000	4464	5 mins ago

Suggest:

1/ RESULT = ALL CASE  - SAME CASE

2/ ALL CASE  = n! (n is length string)

3/ SAME CASE = a! * b! * ... * c! (with a,b,c is number of same chars ) 

from functools import cache

@cache

def fac(n):

    if n == 1:

        return 1

    return fac(n - 1) * n

n = int(input())

for i in range(1, n+1):

    s = input()

   

    d = {}

    for c in s:

        if c not in d:

            d[c] = 1

        else:

            d[c] += 1

    r = fac(len(s))

    for key in d:

        r //= fac(d[key])

    print(f'Data set {i}: {r}')

Solution UVA: 10268 - 498-bis

Problem	Verdict	Lang	Time	Best	Rank	Submit Time
| discuss10268 - 498-bis	Accepted	C++11	0.020	0.000	705	1 mins ago

Susgest:

- Use pow default function couldn't AC so we need code Pow binary function  

#include <bits/stdc++.h>

#define int long long

using namespace std;

string line1, line2;

long long Pow(long long a, long long b) {

    if (!b) return 1;

    long long x = Pow(a, b / 2);

    if (b % 2 == 0)

        return x * x;

    else

        return x * x * a;

}

int32_t main() {

    ios::sync_with_stdio(false);

    cin.tie(nullptr);

    while(getline(cin, line1)){

        getline(cin, line2);

               

        stringstream l1(line1);

        stringstream l2(line2);

        int x;  l1 >> x;

       

        vector<int> a;

        int v;

        while(l2 >> v) a.push_back(v);  

        reverse(a.begin(), a.end());

        int s = 0;

        for(int i=1; i<a.size(); i++)

            s += a[i]*i*Pow(x, i-1);

        cout<<s<<"\n";

    }

}