P1965 夜夜的数据加强 题解

这是VIJIOS上的一道题，原题的链接https://vijos.org/p/1965

先上代码，用C++写的

# include<iostream>using namespace std;int fa[100001];int savemod[400][400][400];int ii[100000];   // a^(i-1)  从1开始使用int jj[100000];   // 1+a+a^2+a^(i-2) 从1开始使用bool ans[400][400][400];int len;int amod[400][400];   // amod[i][j] = i % j;int main(){    int x, a, b;    int i, j, k;    int n, c;    cin >> n >> c;    for (i = 1; i <= n-1; i++)    {        cin >> fa[i];        if (fa[i] >= i || fa[i] < 0 || fa[i] >= c)        {            cout << "Unsuccessful Hack Attempt" << endl;            return 0;        }    }    if (n == 1)    {        for (x = 0; x < c; x++)            for (a = 0; a < c; a++)                for (b = 0; b < c; b++)                    cout << x << " " << a << " " << b << endl;        return 0;    }    int tmp;    for (i = 0; i < c; i++)        for (j = i; j < c; j++)        {            tmp = (i*j) % c;            for (k = 0; k < c - tmp; k++)             {                savemod[i][j][k] = tmp + k;                savemod[j][i][k] = tmp + k;            }            for (k = 0; k < tmp; k++)            {                savemod[i][j][c-tmp+k] = k;                savemod[j][i][c-tmp+k] = k;            }        }        for (i = 1; i < 400; i++)            for (j = 2; j < 400; j++) amod[i][j] = i % j;        int s;        bool flag = false;        if (n - 1 > c)        {            for (a = 0; a < c; a++)            {                for (b = 0; b < c; b++)                {                    for (i = n-2; i >= c; i--)                    {                        if (savemod[fa[i]][a][b] != fa[i+1]) break;                    }                    if (i < c)                         for (x = 0; x < c; x++)                        {                            for (s = savemod[x][a][b],j = 2; j <= c && amod[s][j] == fa[j]; s=savemod[s][a][b], j++);                            if (j > c)                            {                                ans[x][a][b] = true;                                flag = true;                            }                        }                }            }        } else        {            for (a = 0; a < c; a++)            {                ii[1] = 1;                for (i = 2; i < n; i++)                {                    ii[i] = savemod[ii[i-1]][a][0];                }                jj[1] = 0;                for (i = 2; i < n; i++)                {                    jj[i] = savemod[jj[i-1]][a][1];                }                for (b = 0; b < c; b++)                    for (x = 0; x < c; x++)                        if (amod[savemod[x][ii[n-1]][savemod[b][jj[n-1]][0]]][n-1] == fa[n-1] && amod[savemod[x][ii[2]][savemod[b][jj[2]][0]]][2] == fa[2])                        {                            for (j = n-2; j>2 && amod[savemod[x][ii[j]][savemod[b][jj[j]][0]]][j] == fa[j]; --j);                            if (j == 2)                            {                                flag = true;                                ans[x][a][b] = true;                            }                        }            }        }        if (flag == false)        {            cout << "Unsuccessful Hack Attempt" << endl;        } else        {            for (i = 0; i < c; i++)                for (j = 0; j < c; j++)                    for (k = 0; k < c; k++)                         if (ans[i][j][k]) cout << i << " " << j << " " << k << endl;        }        return 0;}

这里参考了该题doc写的题解，题解的地址为https://vijos.org/p/1965/solution。如果单纯用doc的暴力方法是不能通过的（能通过第5个测试点，但是后面的测试点不能通过），我另外加了一个优化。

考虑到N比C大很多的情况，这个时候计算fa[i]时取余运算（%i），就不起作用了，可以简化为
fa[i+1] = (fa[i] * a + b) % c（这里i >= c）
在这里只和a和b有关，与x无关，因此可以先枚举a和b，然后检验a和b是否可行，如果可行就进行，接下来枚举x再校验数列i<=c的情况（此时后面不需要校验了）。

另外这个程序要仔细编写，做尽可能的预处理，下面的测试结果可以看到在第5个测试点是很惊险地通过了。

测试数据 #0: Accepted, time = 15 ms, mem = 315184 KiB, score = 5测试数据 #1: Accepted, time = 0 ms, mem = 315180 KiB, score = 5测试数据 #2: Accepted, time = 0 ms, mem = 315180 KiB, score = 5测试数据 #3: Accepted, time = 0 ms, mem = 315184 KiB, score = 5测试数据 #4: Accepted, time = 937 ms, mem = 315184 KiB, score = 5测试数据 #5: Accepted, time = 0 ms, mem = 315184 KiB, score = 5测试数据 #6: Accepted, time = 0 ms, mem = 315180 KiB, score = 5测试数据 #7: Accepted, time = 0 ms, mem = 315180 KiB, score = 5测试数据 #8: Accepted, time = 15 ms, mem = 315188 KiB, score = 5测试数据 #9: Accepted, time = 0 ms, mem = 315180 KiB, score = 5测试数据 #10: Accepted, time = 78 ms, mem = 315184 KiB, score = 5测试数据 #11: Accepted, time = 93 ms, mem = 315184 KiB, score = 5测试数据 #12: Accepted, time = 93 ms, mem = 315184 KiB, score = 5测试数据 #13: Accepted, time = 78 ms, mem = 315184 KiB, score = 5测试数据 #14: Accepted, time = 93 ms, mem = 315180 KiB, score = 5测试数据 #15: Accepted, time = 390 ms, mem = 315180 KiB, score = 5测试数据 #16: Accepted, time = 296 ms, mem = 315180 KiB, score = 5测试数据 #17: Accepted, time = 375 ms, mem = 315180 KiB, score = 5测试数据 #18: Accepted, time = 437 ms, mem = 315184 KiB, score = 5测试数据 #19: Accepted, time = 359 ms, mem = 315184 KiB, score = 5Accepted, time = 3259 ms, mem = 315188 KiB, score = 100