p152.ans1
======================
50 50

=================
p152.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<int> a;

void read() {
    cin >> n;
    a.resize(n);
    cin >> a;
}

void solve() {
    // Each candidate's real share is a[i]*100/sum percent. We first floor
    // every share, which leaves a remainder sum_p = 100 - sum of floors to
    // distribute (it never exceeds the number of candidates with a fractional
    // part). We then walk the list and round up exactly the candidates whose
    // share was not already an integer, one each, until sum_p is exhausted.
    // Every printed part stays within one of its real value and the total is
    // exactly 100.

    int sum = 0;
    for(int x: a) {
        sum += x;
    }

    vector<int> p(n);
    int sum_p = 100;
    for(int i = 0; i < n; i++) {
        p[i] = (a[i] * 100) / sum;
        sum_p -= p[i];
    }

    for(int i = 0; i < n; i++) {
        if(p[i] * 1ll * sum != a[i] * 100ll && sum_p) {
            p[i]++;
            sum_p--;
        }
    }

    cout << p << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}

=================
p152.in1
======================
2
10 10

=================
statement.txt
======================
152. Making round
time limit per test: 0.25 sec.
memory limit per test: 4096 KB
input: standard input
output: standard output



Extraordinary things are around us! All citizens of democratic Berland provided election of president and nobody voted "against all candidates". All votes distributed between N candidates. You was charged to write a program to calculate the results of election. You must indicate a part of votes given to each of the candidates. You must write integer numbers, concidering two conditions:
- Total sum of parts must be exactly 100%
- Every part is an integer number and it must be equal to real part, rounded up or down.

Input
There is a natural number N (1<=N<=10000) written in the first line - amount of candidates. Second line contains sequence of non-negative integer numbers A1, A2,..., AN; Ai is amount of votes given to the i-th candidate (Ai<=10000). Numbers of sequence are separated by one or more spaces.

Output
Write sequence of sought parts. Separate numbers by a space. If solution does not exist, write "No solution". If there are several solutions write any of them.

Sample test(s)

Input
2
10 10

Output
50 50
Author:	Michael R. Mirzayanov
Resource:	Saratov Subregional School Team Contest, 2002
Date:	Spring, 2002








=================
