uva10328 Coin Toss（dp）

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uva10328

题目

就是说抛硬币正面为H，反面为T，问抛n次的所有可能情况中连续H的个数至少为k的个数。

思路

首先要反一下思路，dp[i][j]记录的是抛i次硬币，连续H的个数不超过j的个数，最后答案就是dp[n][n]-dp[n][k-1],对于dp[i][j]怎么求，首先dp[i][j]=2*dp[i-1][j],相当于强行加上正反面，但是有超过j的情况要排除，如果i-1==j那就减一，不赘述。如果i-1>j,那么我们可以确定的是最后有j个H以及最后j+1是T，不确定的就是1~（i-j-2），而且1~（i-j-2）中H的个数不超过j，所有满足的都要去掉，也就是去掉dp[i-j-2][j],最后就是大数了。
而且循环的时候第二层循环要到n为止，后面的转移会用到。

代码

#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<stack>#include<cstdlib>#include<algorithm>#include<cmath>using namespace std;typedef long long ll;const int maxn=110;const int N = 110;int n,k;struct bign{    int len, sex;    int s[2000];    bign()    {        this -> len = 1;        this -> sex = 0;        memset(s, 0, sizeof(s));    }    bign operator = (const char *number)    {        int begin = 0;        len = 0;        sex = 1;        if (number[begin] == '-')        {            sex = -1;            begin++;        }        else if (number[begin] == '+')            begin++;        for (int j = begin; number[j]; j++)            s[len++] = number[j] - '0';    }    bign operator = (int number)    {        char string[N];        sprintf(string, "%d", number);        *this = string;        return *this;    }    bign (int number)    {        *this = number;    }    bign (const char* number)    {        *this = number;    }    bign change(bign cur)    {        bign now;        now = cur;        for (int i = 0; i < cur.len; i++)            now.s[i] = cur.s[cur.len - i - 1];        return now;    }    void delZore()      // 删除前导0.    {        bign now = change(*this);        while (now.s[now.len - 1] == 0 && now.len > 1)        {            now.len--;        }        *this = change(now);    }    void put()      // 输出数值。    {        delZore();        if (sex < 0 && (len != 1 || s[0] != 0))            cout << "-";        for (int i = 0; i < len; i++)            cout << s[i];    }    bign operator + (const bign &cur)    {        bign sum, a, b;        sum.len = 0;        a = a.change(*this);        b = b.change(cur);        for (int i = 0, g = 0; g || i < a.len || i < b.len; i++)        {            int x = g;            if (i < a.len) x += a.s[i];            if (i < b.len) x += b.s[i];            sum.s[sum.len++] = x % 10;            g = x / 10;        }        return sum.change(sum);    }    bign operator - (const bign &cur)    {        bign sum, a, b;        sum.len = len;        a = a.change(*this);        b = b.change(cur);        for (int i = 0; i < b.len; i++)        {            sum.s[i] = a.s[i] - b.s[i] + sum.s[i];            if (sum.s[i] < 0)            {                sum.s[i] += 10;                sum.s[i + 1]--;            }        }        for (int i = b.len; i < a.len; i++)        {            sum.s[i] += a.s[i];            if (sum.s[i] < 0)            {                sum.s[i] += 10;                sum.s[i + 1]--;            }        }        return sum.change(sum);    }};bign dp[maxn][maxn];int main(){    for(int i=0; i<=100; i++)        dp[i][0]=dp[0][i]=1;    for(int i=1; i<=100; i++)        for(int j=1; j<=100; j++)        {            dp[i][j]=dp[i-1][j]+dp[i-1][j];            if(i==j+1)                dp[i][j]=dp[i][j]-bign(1);            else if(i>j+1)                dp[i][j]=dp[i][j]-dp[i-j-2][j];        }    while(scanf("%d%d",&n,&k)!=EOF)    {        bign temp=dp[n][n]-dp[n][k-1];        temp.put();        printf("\n");    }    return 0;}

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