UVA 10328 - Coin Toss dp+大数

题目链接：

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1269

10328 - Coin Toss

Time limit: 3.000 seconds #### 问题描述 > 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. > In the above example we see 1 sequence 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 2n times. How many sequence you will get which contains a sequence of H of > length at least k.

输入

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

输出

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

样例输入

4 1
4 2
4 3
4 4
6 2

样例输出

15
8
3
1
43

题意

给你n个硬币，考虑正反的所有排列数：比如n=2:{HH,HT,TH,TT)，然后问至少有k个的硬币连续正面朝上的总数。

题解

至少k个=2^n-最多k-1个。
所以我们可以转换成去求最多k个的问题
dp[k][i][0]表示前i个最多k个连续正面朝上，且第i个反面朝上的总数，
dp[k][i][1]表示前i个最多k个连续正面朝上，且第i个反面朝上的总数，
则我们只需要考虑扣掉最后连续k+1个正面朝上这种情况就可以转移了，写完之后上大整数。

代码

c++(没考虑数据溢出):

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=111;

LL dp[maxn][maxn][2];

int n,m;

void pre(){
    clr(dp,0);
    for(int k=0;k<maxn;k++){
        dp[k][0][0]=1;
        for(int i=1;i<maxn;i++){
            dp[k][i][1]=dp[k][i][0]=dp[k][i-1][0]+dp[k][i-1][1];
            if(i>k) dp[k][i][1]-=dp[k][i-k-1][0];
        }
    }
}


int main() {
    pre();
    while(scf("%d%d",&n,&m)==2){
        LL x=1;
        rep(i,0,n) x*=2;
        prf("%lld\n",x-dp[m-1][n][0]-dp[m-1][n][1]);
    }
    return 0;
}

//end-----------------------------------------------------------------------

java:

import java.util.*;
import java.math.*;

public class Main {
	final static int maxn = 111;

	public static void main(String args[]) {
		Scanner cin = new Scanner(System.in);
		BigInteger[][][] dp = new BigInteger[maxn][maxn][2];

		for (int i = 0; i < maxn; i++) {
			for (int j = 0; j < maxn; j++) {
				dp[i][j][1]=dp[i][j][0] = BigInteger.ZERO;
			}
		}

		for (int k = 0; k < maxn; k++) {
			dp[k][0][0] = BigInteger.ONE;
			for (int i = 1; i < maxn; i++) {
				dp[k][i][1]=dp[k][i][0] = dp[k][i - 1][0].add(dp[k][i - 1][1]);
				
				if (i > k)
					dp[k][i][1] = dp[k][i][1].subtract(dp[k][i - k - 1][0]);
			}
		}

		BigInteger[] x = new BigInteger[maxn];
		x[0] = BigInteger.ONE;
		for (int i = 1; i < maxn; i++)
			x[i] = x[i - 1].add(x[i - 1]);

		while (cin.hasNext()) {
			int n, m;
			n = cin.nextInt();
			m = cin.nextInt();

			BigInteger ans = x[n].subtract(dp[m - 1][n][0]).subtract(dp[m - 1][n][1]);

			System.out.println(ans.toString());

		}
	}
}