题意：有N堆石子（N为大数），每堆的个数按一定方式生成，问先手取若干堆进行尼姆博弈，必胜的方式有多少种。
题解：因为 k < 4096，所以出现的数最多只有4096个，对每个数字只考虑出现奇/偶次进行dp，答案是所有不等于0的dp值的和。然后如果一个数字出现x次，它对自己出现奇数次的方案数和偶数次的方案数贡献都是乘上2 ^ (x - 1)，每个数字的贡献都是2 ^ (x - 1) 总贡献就是2 ^ (N - ∑(vis[i]))。大数可以一边读入一边取模。
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4 #define ull unsigned long long
5 #define mst(a,b) memset((a),(b),sizeof(a))
6 #define mp(a,b) make_pair(a,b)
7 #define fi first
8 #define se second
9 #define pi acos(-1)
10 #define pii pair<int,int>
11 const int INF = 0x3f3f3f3f;
12 const double eps = 1e-6;
13 const int MAXN = 1e7 + 10;
14 const int MAXM = 2e6 + 10;
15 const ll mod = 1e9 + 7;
16
17 bool vis[4105];
18 int dp[2][8192];
19 vector<int>vec;
20 int f[4105];
21
22 ll pow_mod(ll a, ll n) {
23 ll ans = 1;
24 while(n) {
25 if(n & 1) ans = ans * a % mod;
26 a = a * a % mod;
27 n >>= 1;
28 }
29 return ans;
30 }
31
32 int main() {
33 #ifdef local
34 freopen("data.txt", "r", stdin);
35 // freopen("data.txt", "w", stdout);
36 #endif
37 ll ans = 0;
38 char ch;
39 int len = 0;
40 int n = 1e9;
41 while(1) {
42 scanf("%c", &ch);
43 if(ch == ‘ ‘) break;
44 ans = pow_mod(ans, 10);
45 if(len == 0) ans = 1;
46 ans = ans * (1ll << (ch - ‘0‘)) % mod;
47 if(len <= 5) {
48 if(len == 0)
49 n = 0;
50 n = n * 10 + ch - ‘0‘;
51 len++;
52 }
53 }
54 int x;
55 scanf("%d", &x);
56 int temp = x;
57 int a, b, c, d, e, k;
58 scanf("%d%d%d%d%d%d", &a, &b, &c, &d, &e, &k);
59 vec.push_back(x);
60 vis[x] = true;
61 for(int i = 1; i <= 4100; i++) {
62 int x1 = 1ll * a * i * i % k * i * i % k;
63 int x2 = 1ll * b * i * i % k * i % k;
64 int x3 = 1ll * c * i * i % k;
65 int x4 = 1ll * d * i;
66 x = (x1 + x2 + x3 + x4 + e - 1) % k + 1;
67 f[i] = x;
68 }
69 int all = 0;
70 while(all < n) {
71 all++;
72 int now = f[temp];
73 if(!vis[now]) {
74 vis[now] = true;
75 vec.push_back(now);
76 temp = now;
77 } else break;
78 }
79 int sz = vec.size();
80 dp[0][0] = 1;
81 int pre = 1, now = 0;
82 for(int i = 1; i <= sz; i++) {
83 swap(pre, now);
84 int num = vec[i - 1];
85 for(int j = 0; j < 8192; j++) dp[now][j] = dp[pre][j];
86 for(int j = 0; j < 8192; j++) {
87 dp[now][j ^ num] += dp[pre][j];
88 if(dp[now][j ^ num] >= mod) dp[now][j ^ num] -= mod;
89 }
90 }
91 ll sum = 0;
92 for(int i = 1; i < 8192; i++) {
93 sum += dp[now][i];
94 if(sum >= mod) sum -= mod;
95 }
96 ll inv = pow_mod(2, sz);
97 inv = pow_mod(inv, mod - 2);
98 ans = ans * inv % mod;
99 ans = ans * sum % mod;
100 printf("%lld\n", ans);
101 return 0;
102 }
原文地址：https://www.cnblogs.com/scaulok/p/9573791.html
时间： 2024-11-06 07:10:52