Tribles
Problem's Link: http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=33059
Mean:
有k个细菌，每个细菌只能存活一天，在死去之前可能会分裂出0,1,2....n-1个细菌，对应的概率为p0,p1,p2....pn-1。
问：所有细菌在第m天全部灭亡的概率是多少？（m天以前灭亡也算在内）
analyse:
由于每一个细菌的生存是独立的，所以我们可以先算出一个细菌的概率为PP，最终答案应是：PP^k。
设dp[i]表示第i天全部灭亡的概率，那么：
dp[i] = p0*(dp[i-1]^0) + p1*(dp[i-1]^1) + p2*(dp[i-1]^2) + ...pn-1*(dp[i-1]^(n-1))
其中pi*(dp[j-1]^i)表示：该细菌分裂成了i个，这i个细菌在第j-1天灭亡的概率。
由于每个细菌独立，所以是乘法，也就是i次方。
对于dp[0]，代表第0天就全部灭亡，也就是根本没有分裂，所以dp[0]=p0.
Time complexity: O(N)
Source code:
/*
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-26-20.36
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
const int MAXN = 1010;
double p [ MAXN ], dp [ MAXN ];
int main()
{
ios_base :: sync_with_stdio( false);
cin . tie( 0);
int t;
scanf( "%d" , & t);
for( int Cas = 1; Cas <= t; ++ Cas)
{
int n , k , m;
scanf( "%d %d %d" , &n , & k , & m);
for( int i = 0; i <n; ++ i)
scanf( "%lf" , &p [ i ]);
dp [ 0 ] =p [ 0 ];
for( int i = 1; i < m; ++ i)
{
dp [ i ] = 0.;
for( int j = 0; j <n; ++ j)
{
dp [ i ] +=p [ j ] * pow( dp [ i - 1 ], j);
}
}
printf( "Case #%d: %.7f \n " , Cas , pow( dp [ m - 1 ], k));
}
return 0;
}
/*
*/
* this code is made by crazyacking
* Verdict: Accepted
* Submission Date: 2015-08-26-20.36
* Time: 0MS
* Memory: 137KB
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
const int MAXN = 1010;
double p [ MAXN ], dp [ MAXN ];
int main()
{
ios_base :: sync_with_stdio( false);
cin . tie( 0);
int t;
scanf( "%d" , & t);
for( int Cas = 1; Cas <= t; ++ Cas)
{
int n , k , m;
scanf( "%d %d %d" , &n , & k , & m);
for( int i = 0; i <n; ++ i)
scanf( "%lf" , &p [ i ]);
dp [ 0 ] =p [ 0 ];
for( int i = 1; i < m; ++ i)
{
dp [ i ] = 0.;
for( int j = 0; j <n; ++ j)
{
dp [ i ] +=p [ j ] * pow( dp [ i - 1 ], j);
}
}
printf( "Case #%d: %.7f \n " , Cas , pow( dp [ m - 1 ], k));
}
return 0;
}
/*
*/
