這道題目其實就是說有N張紙牌，問至少連續K張正面朝上的可能性是多少。

可以用遞推做。
首先我們將題目所求從

至少K張

轉化為

總數 - 至多K張

（為什么要這樣自己想）

設F[i][j]為前i個紙牌至多K張的種數。其中j記錄第i張紙牌的狀態，1為正面朝上，0為反面。

那么可以總結出

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

1 f[i][1] = f[i - 1][0] + f[i - 1][1]; 2 if(i == k) f[i][1] -= 1; 3 else if(i > k) f[i][1] -= f[i - k][0];

最后要記住這道題需要寫高精，我第一次就是這么撲街的。

注：高精用的模板，手殘就全加上了，事實上只用加減法。

AC代碼如下（洛谷上不知道為什么被積極拒絕，只好去Vjudge上提交。。。沒去UVa才不是因為我沒有注冊。）

1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstring> 5 #include <cmath> 6 #include <vector> 7 #include <cassert> 8 using namespace std; 9 10 const int maxn = 105; 11 12 struct BigInteger { 13 typedef unsigned long long LL; 14 15 static const int BASE = 100000000; 16 static const int WIDTH = 8; 17 vector<int> s; 18 19 BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;} 20 BigInteger(LL num = 0) {*this = num;} 21 BigInteger(string s) {*this = s;} 22 BigInteger& operator = (long long num) { 23 s.clear(); 24 do { 25 s.push_back(num % BASE); 26 num /= BASE; 27 } while (num > 0); 28 return *this; 29 } 30 BigInteger& operator = (const string& str) { 31 s.clear(); 32 int x, len = (str.length() - 1) / WIDTH + 1; 33 for (int i = 0; i < len; i++) { 34 int end = str.length() - i*WIDTH; 35 int start = max(0, end - WIDTH); 36 sscanf(str.substr(start,end-start).c_str(), "%d", &x); 37 s.push_back(x); 38 } 39 return (*this).clean(); 40 } 41 42 BigInteger operator + (const BigInteger& b) const { 43 BigInteger c; c.s.clear(); 44 for (int i = 0, g = 0; ; i++) { 45 if (g == 0 && i >= s.size() && i >= b.s.size()) break; 46 int x = g; 47 if (i < s.size()) x += s[i]; 48 if (i < b.s.size()) x += b.s[i]; 49 c.s.push_back(x % BASE); 50 g = x / BASE; 51 } 52 return c; 53 } 54 BigInteger operator - (const BigInteger& b) const { 55 assert(b <= *this); // 減數不能大于被減數 56 BigInteger c; c.s.clear(); 57 for (int i = 0, g = 0; ; i++) { 58 if (g == 0 && i >= s.size() && i >= b.s.size()) break; 59 int x = s[i] + g; 60 if (i < b.s.size()) x -= b.s[i]; 61 if (x < 0) {g = -1; x += BASE;} else g = 0; 62 c.s.push_back(x); 63 } 64 return c.clean(); 65 } 66 BigInteger operator * (const BigInteger& b) const { 67 int i, j; LL g; 68 vector<LL> v(s.size()+b.s.size(), 0); 69 BigInteger c; c.s.clear(); 70 for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j]; 71 for (i = 0, g = 0; ; i++) { 72 if (g ==0 && i >= v.size()) break; 73 LL x = v[i] + g; 74 c.s.push_back(x % BASE); 75 g = x / BASE; 76 } 77 return c.clean(); 78 } 79 BigInteger operator / (const BigInteger& b) const { 80 assert(b > 0); // 除數必須大于0 81 BigInteger c = *this; // 商:主要是讓c.s和(*this).s的vector一樣大 82 BigInteger m; // 余數:初始化為0 83 for (int i = s.size()-1; i >= 0; i--) { 84 m = m*BASE + s[i]; 85 c.s[i] = bsearch(b, m); 86 m -= b*c.s[i]; 87 } 88 return c.clean(); 89 } 90 BigInteger operator % (const BigInteger& b) const { //方法與除法相同 91 BigInteger c = *this; 92 BigInteger m; 93 for (int i = s.size()-1; i >= 0; i--) { 94 m = m*BASE + s[i]; 95 c.s[i] = bsearch(b, m); 96 m -= b*c.s[i]; 97 } 98 return m; 99 } 100 // 二分法找出滿足bx<=m的最大的x 101 int bsearch(const BigInteger& b, const BigInteger& m) const{ 102 int L = 0, R = BASE-1, x; 103 while (1) { 104 x = (L+R)>>1; 105 if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;} 106 else R = x; 107 } 108 } 109 BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;} 110 BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;} 111 BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;} 112 BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;} 113 BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;} 114 115 bool operator < (const BigInteger& b) const { 116 if (s.size() != b.s.size()) return s.size() < b.s.size(); 117 for (int i = s.size()-1; i >= 0; i--) 118 if (s[i] != b.s[i]) return s[i] < b.s[i]; 119 return false; 120 } 121 bool operator >(const BigInteger& b) const{return b < *this;} 122 bool operator<=(const BigInteger& b) const{return !(b < *this);} 123 bool operator>=(const BigInteger& b) const{return !(*this < b);} 124 bool operator!=(const BigInteger& b) const{return b < *this || *this < b;} 125 bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);} 126 }; 127 128 ostream& operator << (ostream& out, const BigInteger& x) { 129 out << x.s.back(); 130 for (int i = x.s.size()-2; i >= 0; i--) { 131 char buf[20]; 132 sprintf(buf, "%08d", x.s[i]); 133 for (int j = 0; j < strlen(buf); j++) out << buf[j]; 134 } 135 return out; 136 } 137 138 istream& operator >> (istream& in, BigInteger& x) { 139 string s; 140 if (!(in >> s)) return in; 141 x = s; 142 return in; 143 } 144 145 int n, k; 146 BigInteger f[maxn][2]; 147 148 int main() { 149 while(cin >> n >> k) { 150 memset(f, 0, sizeof(f)); 151 f[0][1] = 1; 152 for(int i = 1; i <= n; i++) { 153 f[i][0] = f[i - 1][0] + f[i - 1][1]; 154 f[i][1] = f[i - 1][0] + f[i - 1][1]; 155 if(i == k) f[i][1] = f[i][1] - 1; 156 else if(i > k) f[i][1] = f[i][1] - f[i - k][0]; 157 } 158 BigInteger a = 1, t = 2; 159 for(int i = 0; i < n; i++) a = t * a; 160 cout << a - f[n][1] - f[n][0] << endl; 161 } 162 }

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轉載于:https://www.cnblogs.com/ilverene/p/UVA10328.html

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