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解法1代码：
//577ms #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int maxn = 1e5 + 10; const ll mod = 998244353; ll p[22], q[22], inv[22]; ll sum, ans; int n; ll q_pow(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res = res * a % mod; a = a * a % mod; b /= 2; } return res; } void dfs(int pos, ll cnt) { if (pos == n) { ans = (ans + cnt) % mod; return; } dfs(pos + 1, cnt); dfs(pos + 1, cnt * q[pos] % mod * (p[pos] - 1) % mod * inv[pos] % mod); } int main(void) { #ifdef ACM freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif int t; scanf("%d", &t); while (t--) { sum = 1; ans = 0; scanf("%d", &n); for (int i = 0; i < n; i++) { scanf("%lld%lld", p + i, q + i); inv[i] = q_pow(p[i], mod - 2); sum = sum * q_pow(p[i], q[i]) % mod; } dfs(0, sum); printf("%lld\n", ans); } return 0; }
//15ms #include #include #include #include #include #include #include #include #include #include #include using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int maxn = 1e5 + 10; const ll mod = 998244353; ll p[22], q[22], inv[22]; ll sum; ll q_pow(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res = res * a % mod; a = a * a % mod; b /= 2; } return res; } int main(void) { #ifdef ACM freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif int t, m; scanf("%d", &t); while (t--) { sum = 1; scanf("%d", &m); for (int i = 0; i < m; i++) { scanf("%lld%lld", p + i, q + i); inv[i] = q_pow(p[i], mod - 2); sum = sum * q_pow(p[i], q[i]) % mod; } for (int i = 0; i < m; i++) { sum = sum * (((q[i] * (p[i] - 1) % mod) * inv[i] + 1) % mod) % mod; } printf("%lld\n", sum % mod); } return 0; }
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