Three Integers

Three Integers

枚举\(a,b\)

处理 \(c\) 最小上界 \(cc = cb * (c / cb) + 1 * cb;\)

最大上界\(cc = cb * (c / cb) + 0 * cb;\)

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define endl '\n'
typedef double db;
typedef long long LL;
const db EPS = 1e-9;
const int N = 2e5 + 100,INF = 1 << 31 - 1;
int main() {
    ios::sync_with_stdio(0),cout.tie(0),cin.tie(0);
    int t,a,b,c;
    cin >> t;
    while(t --) {
        cin >> a >> b >> c;
        int x,y,z,ans = INF;
        for(int ca = 1;ca <= 2 * a; ++ca) {
            for(int cb = ca;cb <= 2 * b; cb += ca) {
                for(int k = 0;k < 2; ++k) {
                    int cc = cb * (c / cb) + k * cb;
                    int ts = abs(a - ca) + abs(b - cb) + abs(c - cc);
                    if(ts < ans) {
                        ans = ts;
                        x = ca,y = cb,z = cc;
                    }
                }
            }
        }
        cout << ans << endl << x <<' ' << y <<' ' << z << endl;
    }
    return 0;
}