1512E - Permutation by Sum - CodeForces Solution


brute force greedy math *1600

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Python Code:

t = int(input())
for tidx in range(t):
    n, l, r, s = [int(x) for x in input().split()]
    if s > n*(n+1)//2 - (n-r+l)*(n-r+l-1)//2 or s < (r-l+1)*(r-l+2)//2:
        print(-1)
    else:
        x = list(range(1, r-l+2))
        m = s - sum(x)
        for i in range(r-l+1):
            j = r-l-i
            xjold = x[j]
            x[j] = min(n-i, x[j] + m)
            m = m - x[j] + xjold
        y = list(set(range(1, n+1)).difference(x))
        print(*y[:(l-1)], *x, *y[(l-1):])
        

C++ Code:

#include <bits/stdc++.h>
using namespace std;
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    int tc; cin >> tc;
    while(tc--) {
        int n, l, r, s;
        cin >> n >> l >> r >> s;
        if(s < (r - l + 1) * (r - l + 2) / 2 || s > (r - l + 1) * (2 * n - r + l) / 2) {
            cout << "-1\n";
            continue;
        }
        vector <int> ans(n + 1);  
        int d = (s - (r - l + 1) * (r - l + 2) / 2) / (r - l + 1);
        int extra = (s - (r - l + 1) * (r - l + 2) / 2) % (r - l + 1);

        set <int> not_used;
        for(int i = 1; i <= n; i++) not_used.insert(i); 
        for(int i = l; i <= r; i++) {
            ans[i] = i - l + d + 1;
            if(r - extra < i) ans[i]++;  
            not_used.erase(ans[i]); 
        }
        for(int i = 1; i < l; i++) {
            ans[i] = *not_used.begin();
            not_used.erase(*not_used.begin());
        }
        for(int i = r + 1; i <= n; i++) {
            ans[i] = *not_used.begin();
            not_used.erase(*not_used.begin());
        }
        for(int i = 1; i <= n; i++) cout << ans[i] << " ";
        cout << "\n"; 
    }
    return 0;
}


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