C Need for Pink Slips

C Need for Pink Slips

// Problem: C. Need for Pink Slips// Contest: Codeforces - Codeforces Round #730 (Div. 2)// URL: Memory Limit: 256 MB// Time Limit: 1000 ms// 2022-02-16 17:48:02// // Powered by CP Editor (namespace std;#define rep(i,l,r) for(int i=(l);i<=(r);i++)#define per(i,l,r) for(int i=(l);i>=(r);i--)#define ll long long#define pii pair#define mset(s,t) memset(s,t,sizeof(t))#define mcpy(s,t) memcpy(s,t,sizeof(t))#define fir first#define pb push_back#define sec second#define sortall(x) sort((x).begin(),(x).end())inline int read () { int x = 0, f = 0; char ch = getchar(); while (!isdigit(ch)) f |= (ch=='-'),ch= getchar(); while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar(); return f?-x:x;}template void print(T x) { if (x < 0) putchar('-'), x = -x; if (x >= 10) print(x/10); putchar(x % 10 + '0');}const long double sc = 1e+6;long double exp (int c, int m, int p, int v) { long double ans = p / sc; if (c > 0) { if (c > v) { if (m > 0) ans += (c / sc) * (1 + exp (c - v, m + v / 2, p + v / 2, v)); else ans += (c / sc)* (1 + exp(c - v, 0, p + v, v )); } else { if (m > 0) ans += (c /sc) *(1 + exp(0, m + c / 2, p + c / 2, v)); else ans += (c / sc) * (1 + exp(0, 0, p + c, v)); } } if (m > 0) { if (m > v) { if (c > 0) ans += (m /sc) * (1 + exp(c + v / 2, m - v ,p + v / 2, v)); else ans += (m / sc) * (1 + exp(0, m - v, p + v, v)); } else { if (c > 0) { ans += (m / sc) * (1 + exp(c + m / 2, 0, p + m / 2, v)); } else ans += (m /sc) * (1 + exp(0, 0, p + m, v)); } } return ans;}void solve() { long double a, b ,c, d; cin >> a >> b >> c >> d; int aa = round(a * sc); int bb = round(b *sc); int cc = round (c * sc); int dd = round(d * sc); long double ans = exp(aa, bb, cc, dd); cout << setprecision(12) << fixed << ans <<"\n";}int main () { int t; cin >> t; while (t --) solve(); return 0;}

看范围对于v都>=0.1,所以每次p增加至少0.5，最多执行20次，2^20 1e7可以完成 对于概率的处理，先转换为整数，然后选答案的时候在转回来

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