牛客练习赛35 C题函数的魔法 (Floyd)

最新推荐文章于 2025-08-15 19:58:33 发布

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最新推荐文章于 2025-08-15 19:58:33 发布

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本文链接：https://blog.youkuaiyun.com/qq1013459920/article/details/89854207

博客给出题目链接，涉及Floyd变形，有dis[i][F(i)]=1、dis[i][G(i)]=1、dis[i][i]=0的设定，最终要求解min(dis[F(A)][B],dis[G(A)][B])。

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题目链接：https://ac.nowcoder.com/acm/contest/326/C

Floyd变形,dis[i][F(i)]=1,dis[i][G(i)]=1,dis[i][i]=0

最后求min(dis[F(A)][B],dis[G(A)][B])

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
typedef long long ll;
static const int MAX_N = 235;
static const ll Mod = 233;
static const int N = 105;
static const int INF = 0x3f3f3f3f;
int dis[MAX_N][MAX_N];
int fun1(int x) { return x * x % Mod * (x + 1) % Mod; }
int fun2(int x) { return x * x % Mod * (x - 1) % Mod; }
void solve() {
    memset(dis, INF, sizeof(dis));
    for (int i = 0; i < 233; ++i) {
        dis[i][fun1(i)] = 1;
        dis[i][fun2(i)] = 1;
        dis[i][i] = 0;
    }
    for (int i = 0; i < 233; ++i) {
        for (int j = 0; j < 233; ++j) {
            for (int k = 0; k < 233; ++k) {
                if (dis[j][i] >= INF || dis[i][k] >= INF) continue;
                dis[j][k] = min(dis[j][k], dis[j][i] + dis[i][k]);
            }
        }
    }
}
int main() {
//  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);
    int T;
    solve();
    scanf("%d", &T);
    while (T--) {
        ll A, B;
        scanf("%lld%lld", &A, &B);
        if (A == B) puts("0");            //A,B相等必须首先判断
        else if (B >= 233) puts("-1");    
        else {
            A %= 233;
            int res = INF;
			if(dis[fun1(A)][B] < INF) res = min(res, dis[fun1(A)][B] + 1);
			if(dis[fun2(A)][B] < INF) res = min(res, dis[fun2(A)][B] + 1);
			if(res >= INF) puts("-1");
			else printf("%d\n", res);
        }
    }
    return 0;
}