本来是要做 SDOI2011 的计算器的，但是写好交上去狂 WA 不止。
垃圾 BZOJ 毁我青春（然而是我又蒟又非）
题目
求解离散对数，输入 p, y, z 求 y^x == z mod p 的最小非负 x
样例输入：
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5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111
样例输出：
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0
1
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2
0
3
1
2
0
no solution
no solution
1
9584351
462803587
代码
此题卡 map ，考虑到 POJ 也没有 unordered_map， tr1/unordered_map，
hash_map ，所以说要自己写一个。
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#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long LL;
namespace hash_map {
const int BKT_SIZ = 1 << 16;
const int BKT_MSK = BKT_SIZ - 1;
const int POL_SIZ = 5E5;
int good_hash(LL a) {
return (a * 2654435761) & 0xFFFFFFFF;
}
struct node {
LL k; int v;
node* nxt;
} POOL[POL_SIZ]; int ppos;
node* Head[BKT_SIZ];
void add(LL k, int v) {
int kk = good_hash(k) & BKT_MSK;
node *cur = Head[kk];
while(cur && cur->k != k) {
cur = cur->nxt;
}
if(cur) cur->v = v;
else {
POOL[ppos].nxt = Head[kk];
Head[kk] = &POOL[ppos++];
Head[kk]->k = k, Head[kk]->v = v;
}
}
int check(LL k) {
int kk = good_hash(k) & BKT_MSK;
node* cur = Head[kk];
while(cur && cur->k != k) {
cur = cur->nxt;
}
if(cur) return cur->v;
return -1;
}
void restore() {
ppos = 0;
memset(POOL, 0, sizeof(POOL));
memset(Head, 0, sizeof(Head));
}
};
LL fastpow(LL y, LL z, LL p) {
LL c = z;
LL ret = 1, cur = y % p;
while(c) {
if(c & 1) ret = (ret * cur) % p;
cur = (cur * cur) % p;
c >>= 1;
}
return ret;
}
LL y, z, p;
void case3() { // Pure BSGS, calc discrete logarithm
// Find such x, s.t. y^x == z mod p
y %= p;
if(!y && !z) { printf("1n"); return; }
if(!y) { printf("no solutionn"); return; }
LL m = LL(ceil(sqrt(p)));
for(int i=0, cyj=z%p;i<=m;i++) {
hash_map :: add(cyj, i);
cyj = (cyj * y) % p;
}
LL ym = fastpow(y, m, p);
for(int i=1, cym=ym;i<=m;i++) {
int val;
if((val = hash_map :: check(cym)) != -1) {
printf("%lldn", i * m - val);
return;
}
cym = (cym * ym) % p;
}
printf("no solutionn");
}
int main() {
while(scanf("%lld %lld %lld", &p, &y, &z) != EOF) {
hash_map :: restore();
case3();
}
}
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