POJ3243 Clever Y大步小步法

Posted 2021-05-22 海岛Blog

Clever Y
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 11799 Accepted: 2932

Description

Little Y finds there is a very interesting formula in mathematics:

$$X^{Y}modZ=K$$

Given X, Y, Z, we all know how to figure out K fast. However, given X, Z, K, could you figure out Y fast?

Input

Input data consists of no more than 20 test cases. For each test case, there would be only one line containing 3 integers X, Z, K (0 ≤ X, Z, K ≤ 10⁹).
Input file ends with 3 zeros separated by spaces.

Output

For each test case output one line. Write “No Solution” (without quotes) if you cannot find a feasible Y (0 ≤ Y < Z). Otherwise output the minimum Y you find.

Sample Input

5 58 33
2 4 3
0 0 0

Sample Output

9
No Solution

Source

POJ Monthly–2007.07.08, Guo, Huayang

问题链接：POJ3243 Clever Y
问题简述：（略）
问题分析：大步小步法的模板题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ3243 Clever Y */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

typedef long long LL;
const int MOD =76532;
LL hs[MOD], id[MOD];
int next2[MOD], head[MOD], top;

void Insert(LL x, LL y)
{
    int k = x % MOD;
    hs[top] = x; id[top] = y;
    next2[top] = head[k]; head[k] = top++;
}

LL Find(int x)
{
    int k = x % MOD;
    for (int i = head[k]; ~i; i = next2[i])
        if (x == hs[i]) return id[i];
    return -1;
}

LL BSGS(LL a, LL b, LL n)
{
    if (b == 1) return 0;
    memset(head, -1, sizeof head);
    top = 1;
    int m = sqrt(n);
    LL p = 1, x = 1;
    for (int i = 0; i < m; i++) {Insert(p * b % n, i); p = p * a % n;}
    for (LL i = m, k; ; i += m) {
        if ((k = Find(x = x * p % n)) != -1) return i - k;
        if (i > n) break;
    }
    return -1;
}

int main()
{
    LL x, z, k, ans;
    while (~scanf("%lld%lld%lld", &x, &z, &k) && (x || z || k)) {
        if ((ans = BSGS(x, k, z)) == -1)
            puts("No Solution");
        else
            printf("%lld\\n",ans);
    }

    return 0;
}