题解 UVA10328 【Coin Toss】

这道题目其实就是说有N张纸牌，问至少连续K张正面朝上的可能性是多少。

可以用递推做。
首先我们将题目所求从

至少K张

转化为

总数 - 至多K张

（为什么要这样自己想）

设F[i][j]为前i个纸牌至多K张的种数。其中j记录第i张纸牌的状态，1为正面朝上，0为反面。

那么可以总结出

 f[i][0] = f[i - 1][0] + f[i - 1][1];  

f[i][1] = f[i - 1][0] + f[i - 1][1];
if(i == k) f[i][1] -= 1;
else if(i > k) f[i][1] -= f[i - k][0];

最后要记住这道题需要写高精，我第一次就是这么扑街的。

注：高精用的模板，手残就全加上了，事实上只用加减法。

AC代码如下（洛谷上不知道为什么被积极拒绝，只好去Vjudge上提交。。。没去UVa才不是因为我没有注册。）

#include <iostream> 
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath> 
#include <vector>
#include <cassert>
using namespace std;

const int maxn = 105;

struct BigInteger {
    typedef unsigned long long LL;

    static const int BASE = 100000000;
    static const int WIDTH = 8;
    vector<int> s;

    BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}
    BigInteger(LL num = 0) {*this = num;}
    BigInteger(string s) {*this = s;}
    BigInteger& operator = (long long num) {
        s.clear();
        do {
            s.push_back(num % BASE);
            num /= BASE;
        } while (num > 0);
        return *this;
    }
    BigInteger& operator = (const string& str) {
        s.clear();
        int x, len = (str.length() - 1) / WIDTH + 1;
        for (int i = 0; i < len; i++) {
            int end = str.length() - i*WIDTH;
            int start = max(0, end - WIDTH);
            sscanf(str.substr(start,end-start).c_str(), "%d", &x);
            s.push_back(x);
        }
        return (*this).clean();
    }

    BigInteger operator + (const BigInteger& b) const {
        BigInteger c; c.s.clear();
        for (int i = 0, g = 0; ; i++) {
            if (g == 0 && i >= s.size() && i >= b.s.size()) break;
            int x = g;
            if (i < s.size()) x += s[i];
            if (i < b.s.size()) x += b.s[i];
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c;
    }
    BigInteger operator - (const BigInteger& b) const {
        assert(b <= *this); // 减数不能大于被减数
        BigInteger c; c.s.clear();
        for (int i = 0, g = 0; ; i++) {
            if (g == 0 && i >= s.size() && i >= b.s.size()) break;
            int x = s[i] + g;
            if (i < b.s.size()) x -= b.s[i];
            if (x < 0) {g = -1; x += BASE;} else g = 0;
            c.s.push_back(x);
        }
        return c.clean();
    }
    BigInteger operator * (const BigInteger& b) const {
        int i, j; LL g;
        vector<LL> v(s.size()+b.s.size(), 0);
        BigInteger c; c.s.clear();
        for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];
        for (i = 0, g = 0; ; i++) {
            if (g ==0 && i >= v.size()) break;
            LL x = v[i] + g;
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c.clean();
    }
    BigInteger operator / (const BigInteger& b) const {
        assert(b > 0);  // 除数必须大于0
        BigInteger c = *this;       // 商:主要是让c.s和(*this).s的vector一样大
        BigInteger m;               // 余数:初始化为0
        for (int i = s.size()-1; i >= 0; i--) {
            m = m*BASE + s[i];
            c.s[i] = bsearch(b, m);
            m -= b*c.s[i];
        }
        return c.clean();
    }
    BigInteger operator % (const BigInteger& b) const { //方法与除法相同
        BigInteger c = *this;
        BigInteger m;
        for (int i = s.size()-1; i >= 0; i--) {
            m = m*BASE + s[i];
            c.s[i] = bsearch(b, m);
            m -= b*c.s[i];
        }
        return m;
    }
    // 二分法找出满足bx<=m的最大的x
    int bsearch(const BigInteger& b, const BigInteger& m) const{
        int L = 0, R = BASE-1, x;
        while (1) {
            x = (L+R)>>1;
            if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}
            else R = x;
        }
    }
    BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}
    BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}
    BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}
    BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}
    BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}

    bool operator < (const BigInteger& b) const {
        if (s.size() != b.s.size()) return s.size() < b.s.size();
        for (int i = s.size()-1; i >= 0; i--)
            if (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
    bool operator >(const BigInteger& b) const{return b < *this;}
    bool operator<=(const BigInteger& b) const{return !(b < *this);}
    bool operator>=(const BigInteger& b) const{return !(*this < b);}
    bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}
    bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}
};

ostream& operator << (ostream& out, const BigInteger& x) {
    out << x.s.back();
    for (int i = x.s.size()-2; i >= 0; i--) {
        char buf[20];
        sprintf(buf, "%08d", x.s[i]);
        for (int j = 0; j < strlen(buf); j++) out << buf[j];
    }
    return out;
}

istream& operator >> (istream& in, BigInteger& x) {
    string s;
    if (!(in >> s)) return in;
    x = s;
    return in;
}

int n, k;
BigInteger f[maxn][2];

int main() {
    while(cin >> n >> k) {
        memset(f, 0, sizeof(f));
        f[0][1] = 1;
        for(int i = 1; i <= n; i++) {
            f[i][0] = f[i - 1][0] + f[i - 1][1]; 
            f[i][1] = f[i - 1][0] + f[i - 1][1];
            if(i == k) f[i][1] = f[i][1] - 1;
            else if(i > k) f[i][1] = f[i][1] - f[i - k][0];
        }
        BigInteger a = 1, t = 2;
        for(int i = 0; i < n; i++) a = t * a;
        cout << a - f[n][1] - f[n][0] << endl;
    }
}