XTU1233 Coins

最新推荐文章于 2023-11-25 08:38:58 发布

随心丶而遇

最新推荐文章于 2023-11-25 08:38:58 发布

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分类专栏： ACM—dp ACM-湘潭各OJ刷题专栏文章标签：湘潭 dp

本文链接：https://blog.youkuaiyun.com/d_x_d/article/details/51067684

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本文探讨了在给定数量的银币中，如何通过动态规划算法计算出连续出现指定次数正面朝上的可能性。通过分析银币排列组合的特性，作者详细解释了如何使用状态转移方程解决此问题，并提供了代码实现。

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  Coins 
Accepted : 85
 Submit : 202
Time Limit : 1000 MS
 Memory Limit : 65536 KB
 

 
 CoinsProblem Description:Duoxida buys a bottle of MaiDong from a vending machine and the machine give her n coins back. She places them in a line randomly showing head face or tail face on. And Duoxida wants to know how many situations that m continuous coins head face on among all possible situations. Two situations are considered different if and only if there is at least one position that the coins' faces are different.
InputThe first line contains a integer T(no more than 20) which represents the number of test cases.
In each test case, a line contains two integers n and m.()
OutputFor each test case, output the result modulo  in one line.
Sample Input2
 4 2
 5 2
 
Sample Output8
 19
 
 SourceXTU OnlineJudge

 
 题意：抛n枚银币，连续m次正面朝上的次数为多少。 

 
 分析：可以用dp来写，用一个a[n]存储n枚银币的所有可能数；dp[i][1]表示已经有连续m次正面朝上了，dp[i][0]表示没有。从dp[m][ ]开始更新，因为少于m的时候全为不可能；大于m的时候分已存在dp[i-1][1]*2和刚好存在两种情况dp[i-1-m][0]；所以可得dp[i][1]=dp[i-1][1]*2+dp[i-1-m][0]，dp[i][0]=a[i]-dp[i][1]。 

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define MAXN 1000010

int n,m;
ll dp[MAXN][2];
ll a[MAXN];

int main()
{
    a[0] =  1;
    for(int i=1; i<MAXN; i++)
        a[i] = a[i-1]*2%MOD;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        dp[m][1] = 1;
        dp[m][0] = (a[m]-dp[m][1]+MOD)%MOD;
        for(int i=0; i<m; i++)
            dp[i][0] = a[i];
        for(int i=m+1; i<=n; i++)
        {
            dp[i][1] = (dp[i-1][1]*2+dp[i-1-m][0])%MOD;
            dp[i][0] = (a[i]-dp[i][1]+MOD)%MOD;
        }
        printf("%lld\n",dp[n][1]);
    }
    return 0;
}
</span>