牛客第一场补题

牛客多校1

D. Chocolate

题意：

A、B轮流吃巧克力，当一个人选择一个点吃的时候，会把这个点及其左下的所有全部吃完，谁先吃完谁就输了，给出巧克力的大小，问谁会赢。

思路：

考虑当一个人吃完只剩一行或一列时，那么下一个吃的人就可以控制把最后一块留给这个人，因此当一个人吃完剩一行和一列的时候，就可以控制剩下一行或者剩下一列给谁，因此除非初始巧克力大小只有1*1，否则先手必胜

#include<bits/stdc++.h>
using namespace std;

int main(){
    int x, y;
    cin>>x>>y;
    if(x == 1 && y == 1){
        cout<<"Walk Alone\n";
    }
    else{
        cout<<"Kelin\n";
    }
    return 0;
}

L.Three Permutations

题意：

初始点(1, ,1 , 1)，给出三个排列a, b, c，每次x->a[y], y->b[z], z->c[x]，给出q组询问，每次询问给出x'、y'、z'，问最少经过多少秒到达(x',y',z')，若无法到达，输出-1

思路：

显然每三秒x->abc[x], y->bca[y], z->cab[z]，此时这个置换只和我当前对应的x，y，z有关，因此可以通过这个置换得到循环节大小、在一个循环里到达某一个位置的时间。然后对于每一个询问，枚举从第0秒、第1秒、第2秒开始所能到达的时间，用扩展中国剩余定理即可求得答案。

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

const int maxn = 1e5 + 10;
int n, q;
ll a[maxn], b[maxn], c[maxn];
ll ia[maxn], ib[maxn], ic[maxn];
ll abc[maxn], bca[maxn], cab[maxn];
ll dista[maxn], distb[maxn], distc[maxn];
ll mi[5], ai[5];

ll exgcd(ll a, ll b, ll &x, ll &y){
	if(b == 0){
		x = 1, y = 0;
		return a;
	}

	ll d = exgcd(b, a % b, y, x);
	y -= a/b * x;
	return d;
}

ll excrt(){
    ll x, y;
    ll a1 = ai[1], m1 = mi[1];
    ll ans = 0;
    for(int i = 2; i <= 3; i++){
        ll a2 = ai[i], m2 = mi[i];
        ll a = m1, b = m2, c = (a2 - a1 % m2 + m2) % m2;
        ll d = exgcd(a, b, x, y);
        if( c % d != 0) return -1;
        x = x * (c / d)  % (b / d);
        ans = a1 + x * m1;
        m1 = m2 / d * m1;
        ans = (ans % m1 + m1) % m1;
        a1 = ans;
    }
    return ans;
}


void query(){
	cin>>q;
	while(q--){
		int x, y, z;
		cin>>x>>y>>z;
		ll t1, t2, t3;
		ai[1] = dista[x];
		ai[2] = distb[y];
		ai[3] = distc[z];
		if(ai[1] == -1 || ai[2] == -1 || ai[3] == -1) t1 = -1;
	    else t1 = excrt();

	    ai[1] = dista[ic[z]];
	    ai[2] = distb[ia[x]];
	    ai[3] = distc[ib[y]];
	    if(ai[1] == -1 || ai[2] == -1 || ai[3] == -1) t2 = -1;
	    else t2 = excrt();;

	    ai[1] = dista[ic[ib[y]]];
	    ai[2] = distb[ia[ic[z]]];
	    ai[3] = distc[ib[ia[x]]];
	   	if(ai[1] == -1 || ai[2] == -1 || ai[3] == -1) t3 = -1;
	    else t3 = excrt();

	    ll ans = 1e18;
	    if(t1 != -1) ans = min(ans, 3 * t1);
	    if(t2 != -1) ans = min(ans, 3 * t2 + 1);
	    if(t3 != -1) ans = min(ans, 3 * t3 + 2);
	    if(ans == 1e18) cout<<"-1\n";
	    else cout<<ans<<"\n";
	}
}


void solve(){
	memset(dista, -1, sizeof(dista));
	memset(distb, -1, sizeof(distb));
	memset(distc, -1, sizeof(distc));
	cin>>n;
	for(int i = 1; i <= n; i++) cin>>a[i], ia[a[i]] = i;
	for(int i = 1; i <= n; i++) cin>>b[i], ib[b[i]] = i;
	for(int i = 1; i <= n; i++) cin>>c[i], ic[c[i]] = i;



	for(int i = 1; i <= n; i++) abc[i] = a[b[c[i]]];
	for(int i = 1; i <= n; i++) bca[i] = b[c[a[i]]];
	for(int i = 1; i <= n; i++) cab[i] = c[a[b[i]]];

	ll ta = 0, tb = 0, tc = 0;
	for(ll i = 1; dista[i] == -1; i = abc[i], ++ta) dista[i] = ta;
	for(ll i = 1; distb[i] == -1; i = bca[i], ++tb) distb[i] = tb;
	for(ll i = 1; distc[i] == -1; i = cab[i], ++tc) distc[i] = tc;
	mi[1] = ta, mi[2] = tb, mi[3] = tc;
	query();
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int T = 1;
	//cin>>T;
	while(T--){
		solve();
	}
	return 0;
}

M.Water

题意：

初始有两个空水杯，容量为a, b,然后要通过倒水、转移的操作喝到d容量的水，并且喝也算一次操作，问最少操作几次能喝到d的水。

思路：

显然若ax+by=d无解时输出-1。那么当有解时怎样得到操作最少呢。

当x>=0, y>=0时，ans = 2 * (x + y)；

否则，ans = 2 *abs(x - y) - 1

通过观察一次函数可以发现再原点附近的(x + y)最小。

因此，通过扩展欧几里得求得x, y后先得到原点附近的x, y,再枚举其左右两个点对答案取小。

#include<bits/stdc++.h>
using namespace std;
using ll = long long;

const int maxn = 1e6 + 10;
ll n, m, k;
ll a ,b , c;
ll ans = 1e18;

ll exgcd(ll a, ll b, ll &x, ll &y){
	if(b == 0){
		x = 1;
		y = 0;
		return a;
	}

	ll d = exgcd(b, a % b, y, x);
	y -= a / b * x;
	return d;
}


void calc(ll t, ll x, ll y){
	ll s = x + t * b, r = y - t * a;
	if(s >= 0 && r >= 0){
		ans = min(ans, 2*(s + r));
	}
	else{
		ans = min(ans, 2*(abs(s) + abs(r)) - 1);
	}
}

void solve(){
	cin>>n;
	for(int i = 1; i <= n; i++){
		ans = 1e18;
		cin>>a>>b>>c;
		if(a > b) swap(a, b);
		ll x ,y;
		ll d = exgcd(a, b ,x , y);
		if(c % d != 0){
			cout<<"-1\n";
			continue;
		}
		a /= d, b /= d, c /= d, x *= c, y *= c;
		ll t0 = -x / b;
		for(int t = t0 - 1; t <= t0 + 1; t++){
			calc(t, x, y);
		}

		t0 = y / a;
		for(int t = t0 - 1; t <= t0 + 1; t++){
			calc(t, x, y);
		}

		cout<<ans<<"\n";
	}
	
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int T = 1;
	//cin>>T;
	while(T--){
		solve();
	}
	return 0;
}