标签:scan force int 概率 add lse count() fine scanf
算每个游戏完成的概率加起来就是答案了, 然后组合数前缀暴力转移就完事了。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 2e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;} mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); int n; LL t[N]; LL T, sum[N]; LL bin[N], ibin[N]; LL F[N], Finv[N], inv[N]; int nn, mm; LL now; LL power(LL a, LL b) { LL ans = 1; while(b) { if(b & 1) ans = ans * a % mod; a = a * a % mod; b >>= 1; } return ans; } LL C(int n, int m) { if(n < 0 || n < m) return 0; return F[n] * Finv[m] % mod * Finv[n - m] % mod; } LL calc(int n, int m) { while(nn < n) { add(now, now); sub(now, C(nn, mm)); nn++; } while(mm > m) { sub(now, C(nn, mm)); mm--; } while(mm < m) { mm++; add(now, C(nn, mm)); } return now; } void init() { bin[0] = F[0] = Finv[0] = inv[1] = 1; for(int i = 1; i < N; i++) { bin[i] = bin[i - 1] * 2 % mod; ibin[i] = power(bin[i], mod - 2); } for(int i = 2; i < N; i++) { inv[i] = (mod - mod / i) * inv[mod % i] % mod; } for(int i = 1; i < N; i++) { F[i] = F[i - 1] * i % mod; } for(int i = 1; i < N; i++) { Finv[i] = Finv[i - 1] * inv[i] % mod; } } int main() { init(); scanf("%d%lld", &n, &T); for(int i = 1; i <= n; i++) { scanf("%d", &t[i]); sum[i] = sum[i - 1] + t[i]; } LL ans = 0; now = 1, nn = 0, mm = 0; for(int i = 1; i <= n; i++) { if(sum[i] > T) break; else if(sum[i] + i <= T) { add(ans, 1); } else { LL most = T - sum[i]; add(ans, calc(i, most) * ibin[i] % mod); } } printf("%lld\n", ans); return 0; } /* */
Codeforces 1194F Crossword Expert
标签:scan force int 概率 add lse count() fine scanf
原文地址:https://www.cnblogs.com/CJLHY/p/11188757.html