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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#581315 | #9376. Game | cbbdhz | TL | 0ms | 3540kb | C++20 | 2.5kb | 2024-09-22 11:49:25 | 2024-09-22 11:49:26 |
Judging History
answer
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int mod = 998244353;
int qp(int a, int b, int mod) {
int ans = 1;
while (b) {
if (b & 1) {
ans = (ll)ans * a % mod;
}
b >>= 1;
a = (ll)a * a % mod;
}
return ans;
}
int dfs_iterative(int chou1, int chou2, int p1, int p2) {
stack<tuple<int, int>> st;
map<pair<int, int>, int> dp; // 用来存储状态的结果
st.push({ chou1, chou2 });
dp[{chou1, chou2}] = -1; // -1 表示该状态尚未处理
while (!st.empty()) {
auto [chou1, chou2] = st.top();
// 如果已经计算过了,直接跳过
if (dp[{chou1, chou2}] != -1) {
st.pop();
continue;
}
// 终止条件
if (chou1 <= 0) {
dp[{chou1, chou2}] = 0;
st.pop();
continue;
}
if (chou2 <= 0) {
dp[{chou1, chou2}] = 1;
st.pop();
continue;
}
// 计算概率
int temp = qp(p1 + p2, mod - 2, mod);
int pa = (ll)p1 * temp % mod;
int pb = (ll)p2 * temp % mod;
// 子状态
int nm1 = gcd(chou1, chou2 - chou1);
int nm2 = gcd(chou1 - chou2, chou2);
// 如果子状态还没计算过,先压入栈中
if (dp.find({ chou1 / nm1, (chou2 - chou1) / nm1 }) == dp.end()) {
st.push({ chou1 / nm1, (chou2 - chou1) / nm1 });
dp[{chou1 / nm1, (chou2 - chou1) / nm1}] = -1;
}
if (dp.find({ (chou1 - chou2) / nm2, chou2 / nm2 }) == dp.end()) {
st.push({ (chou1 - chou2) / nm2, chou2 / nm2 });
dp[{(chou1 - chou2) / nm2, chou2 / nm2}] = -1;
}
// 如果所有子状态都处理完毕,处理当前状态
if (dp[{chou1 / nm1, (chou2 - chou1) / nm1}] != -1 &&
dp[{(chou1 - chou2) / nm2, chou2 / nm2}] != -1) {
dp[{chou1, chou2}] = ((ll)pa * dp[{chou1 / nm1, (chou2 - chou1) / nm1}] % mod +
(ll)pb * dp[{(chou1 - chou2) / nm2, chou2 / nm2}] % mod) % mod;
st.pop();
}
}
return dp[{chou1, chou2}];
}
void solve() {
int a, b;
cin >> a >> b;
int temp = gcd(a, b);
a /= temp;
b /= temp;
int x, y, z;
cin >> x >> y >> z;
cout << dfs_iterative(a, b, x, y) << endl;
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
int T;
cin >> T;
while (T--) {
solve();
}
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 0ms
memory: 3540kb
input:
3 1 1 2 2 6 1 3 2 3 6 3 4 7 3 15
output:
499122177 910398850 220911476
result:
ok 3 lines
Test #2:
score: -100
Time Limit Exceeded
input:
100000 1 1000000000 12980050 128257807 266126484 1 1000000000 400255084 123438563 768881284 1000000000 1000000000 24563487 72082135 450057094 1 1000000000 56952077 40876000 193815114 1000000000 1000000000 82048274 239365585 326520865 1000000000 1 309821265 346013425 963168258 1 1 104158269 199365020...