UVA 10328 Coin Toss

Coin Toss

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 10328
64-bit integer IO format: %lld      Java class name: Main

 

Toss is an important part of any event. When everything becomes equal toss is the ultimate decider. Normally a fair coin is used for Toss. A coin has two sides head(H) and tail(T). Superstition may work in case of choosing head or tail. If anyone becomes winner choosing head he always wants to choose head. Nobody believes that his winning chance is 50-50. However in this problem we will deal with a fair coin and n times tossing of such a coin. The result of such a tossing can be represented by a string. Such as if 3 times tossing is used then there are possible 8 outcomes.

HHH HHT HTH HTT THH THT TTH TTT

As the coin is fair we can consider that the probability of each outcome is also equal. For simplicity we can consider that if the same thing is repeated 8 times we can expect to get each possible sequence once.

The Problem

In the above example we see 1 sequnce has 3 consecutive H, 3 sequence has 2 consecutive H and 7 sequence has at least single H. You have to generalize it. Suppose a coin is tossed n times. And the same process is repeated 2^n times. How many sequence you will get which contains a consequnce of H of length at least k.

The Input

The input will start with two positive integer, n and k (1<=k<=n<=100). Input is terminated by EOF.

The Output

For each test case show the result in a line as specified in the problem statement.

Sample Input

4 1
4 2
4 3
4 4
6 2

Sample Output

15
8
3
1
43

解题：解题思路跟zoj 3747 一样

dp[i][0] 表示连续u个正面 且第i个是正面的方案数

需要注意的是 这道题目是需要用大数的，也就是需要高精度

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 #define MAXN 100
 struct HP {
     int len,s[MAXN];
     HP() {
         memset(s,,sizeof(s));
         len=;
     }
     HP operator =(const char *num) { //字符串赋值
         len=strlen(num);
         for(int i=; i<len; i++) s[i]=num[len-i-]-'';
     }

     HP operator =(int num) { //int 赋值
         char s[MAXN];
         sprintf(s,"%d",num);
         *this=s;
         return *this;
     }

     HP(int num) {
         *this=num;
     }

     HP(const char*num) {
         *this=num;
     }

     string str()const { //转化成string
         string res="";
         for(int i=; i<len; i++) res=(char)(s[i]+'')+res;
         if(res=="") res="";
         return res;
     }

     HP operator +(const HP& b) const {
         HP c;
         c.len=;
         for(int i=,g=; g||i<max(len,b.len); i++) {
             int x=g;
             if(i<len) x+=s[i];
             if(i<b.len) x+=b.s[i];
             c.s[c.len++]=x%;
             g=x/;
         }
         return c;
     }
     void clean() {
         while(len >  && !s[len-]) len--;
     }

     HP operator *(const HP& b) {
         HP c;
         c.len=len+b.len;
         for(int i=; i<len; i++)
             for(int j=; j<b.len; j++)
                 c.s[i+j]+=s[i]*b.s[j];
         for(int i=; i<c.len-; i++) {
             c.s[i+]+=c.s[i]/;
             c.s[i]%=;
         }
         c.clean();
         return c;
     }

     HP operator - (const HP& b) {
         HP c;
         c.len = ;
         for(int i=,g=; i<len; i++) {
             int x=s[i]-g;
             if(i<b.len) x-=b.s[i];
             if(x>=) g=;
             else {
                 g=;
                 x+=;
             }
             c.s[c.len++]=x;
         }
         c.clean();
         return c;
     }
     HP operator / (const HP &b) {
         HP c, f = ;
         for(int i = len-; i >= ; i--) {
             f = f*;
             f.s[] = s[i];
             while(f>=b) {
                 f =f-b;
                 c.s[i]++;
             }
         }
         c.len = len;
         c.clean();
         return c;
     }
     HP operator % (const HP &b) {
         HP r = *this / b;
         r = *this - r*b;
         return r;
     }

     HP operator /= (const HP &b) {
         *this  = *this / b;
         return *this;
     }

     HP operator %= (const HP &b) {
         *this = *this % b;
         return *this;
     }

     bool operator < (const HP& b) const {
         if(len != b.len) return len < b.len;
         for(int i = len-; i >= ; i--)
             if(s[i] != b.s[i]) return s[i] < b.s[i];
         return false;
     }

     bool operator > (const HP& b) const {
         return b < *this;
     }

     bool operator <= (const HP& b) {
         return !(b < *this);
     }

     bool operator == (const HP& b) {
         return !(b < *this) && !(*this < b);
     }
     bool operator != (const HP &b) {
         return !(*this == b);
     }
     HP operator += (const HP& b) {
         *this = *this + b;
         return *this;
     }
     bool operator >= (const HP &b) {
         return *this > b || *this == b;
     }

 };

 istream& operator >>(istream &in, HP& x) {
     string s;
     in >> s;
     x = s.c_str();
     return in;
 }

 ostream& operator <<(ostream &out, const HP& x) {
     out << x.str();
     return out;
 }
 const int maxn = ;
 HP dp[maxn][];//dp[i][0]表示第i个正
 int n,k;
 HP solve(int u){
     dp[][] = ;
     dp[][] = ;
     for(int i = ; i <= n; ++i){
         if(i <= u) dp[i][] = dp[i-][] + dp[i-][];
         if(i == u + ) dp[i][] = dp[i-][] + dp[i-][] - ;
         if(i > u + ) dp[i][] = dp[i-][] + dp[i-][] - dp[i - u - ][];
         dp[i][] = dp[i-][] + dp[i-][];
     }
     return (dp[n][] + dp[n][]);
 }
 int main(){
     while(~scanf("%d%d",&n,&k))
         cout<<solve(n) - solve(k-)<<endl;
     return ;
 }