题解 | #编程题1#

思路

枚举d1和d2的四个组合，由于代码存在重复性，所以可以写成一个函数进行检验。
题目不难重在理清思路

代码

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using vi = vector<int>;
using vl = vector<ll>;
using vvl = vector< vector<ll> >;
using vvi = vector< vector<int> >;
using vvvi = vector<vector< vector<int> >>;
const ll mod = 1e9 + 7;
const int maxn = 1e7 + 5;
template <typename T1, typename T2> inline void chmin(T1& x, T2 y) { if (x > y) x = y; }
template <typename T1, typename T2> inline void swapmin(T1& x, T2& y) { if (x > y) swap(x, y); }
template <typename T1, typename T2> inline void chmax(T1& x, T2 y) { if (x < y) x = y; }
template <typename T1, typename T2> inline void swapmax(T1& x, T2& y) { if (x < y) swap(x, y); }

ll n, k;

bool check(ll d1, ll d2)
{
    if(n % 3 != 0 || (k - d1 - d2) % 3 != 0)
        return false;
    ll s1, s2, s3, delta;
    s2 = (k - d1 - d2) / 3;
    s1 = s2 + d1;
    s3 = s2 + d2;
    if(s1 > k || s1 < 0 || s2 > k || s2 < 0 || s3 > k || s3 < 0)
        return false;
    swapmax(s1, s2);
    swapmax(s2, s3);//此时s1 >= s2 >= s3
    delta = s1 - s2 + s1 -s3;//要弥补差距的最小分数
    if(n - k >= delta && (n-k - delta) % 3 == 0)
        return true;
    return false;
}

int main()
{
    ll q = 1,  d1, d2;
    cin.tie(0);
    ios::sync_with_stdio(0);
    cin >> q;
    while (q--)
    {
        cin >> n >> k >> d1 >> d2;
        if(check(d1,d2) || check(-d1,-d2) || check(d1,-d2) || check(-d1,d2))
            printf("yes\n");
        else
            printf("no\n");
    }
}