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  • Codeforces 1194F Crossword Expert

    Crossword Expert

    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
    ", ans);
        return 0;
    }
    
    /*
    
    */
    

      

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  • 原文地址:https://www.cnblogs.com/CJLHY/p/11188757.html
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