CF 1379 B. Dubious Cyrpto

传送门

题目：给定一个区间范围[l, r]和m，在区间[l, r]中取a, b, c，使得 na + b - c = m（n > 0）。

思路：na + b - c = m，可以看出a和n都是整数,则：

na + b - c = m -> na = m - b + c

b,c∈[l, r] ->  na ∈ [m + l - r, m + r - l]

则我们需要找到一个na存在于区间[m + l - r, m + r - l]，a∈[l, r]，我们可以枚举a，对于该a：

如果：①m / a * a >= m + l - r 且 m / a * a 一定满足小于等于m，则我们把m - (m / a) * a用b,c∈[l, r]凑出来。

否则：②(m / a  + 1) * a <= m + r - l 且 (m / a + 1) * a一定满足大于等于m，则我们把m - (m / a + 1) * a用b,c∈[l, r]凑出来

当然有一种特殊情况：③ m < l,则直接 a = l, c = r, b = m - l + r

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
 
using namespace std;
 
#define ll long long
#define pb push_back
#define fi first
#define se second

const int N = 2e5 + 10;
char s[N];
int len;

void solve()
{   
    int T;
    cin >> T;
    while(T--){
        
        ll l, r, m, a, b, c;
        cin >> l >> r >> m;
        
        if(m < l){
            a = l;
            c = r;
            b = m + r - l;
            cout << a << " " << b << " " << c << endl;
            continue;
        }

        ll ml = m + l - r;
        ll mr = m + r - l;
        for(int i = l; i <= r; ++i){
            ll times = m / i;
            if(times * i < ml){
                if((times + 1) * i <= mr){
                    a = i;
                    ll remains = (times + 1) * i - m;
                    b = l;
                    c = l + remains;
                    break;
                }
            }else if(times * i >= ml){
                ll remains = m - (times * i);
                a = i;
                b = r;
                c = r - remains;
                break;
            }
        }

        cout << a << " " << b << " " << c << endl;
    }
}
 
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0); 
    solve();
 
    return 0;
}