[Ybt gold medal navigation 1-1-9] guard challenge

Guard challenge

Topic link: ybt gold medal navigation 1-1-9

Topic

There are some tasks, and each task has a certain chance of success.
Success will add a value to a number initially K. (Different tasks will subtract different values, and there will be -1 and positive integers.)
Then you are asked to ensure that the probability of this value being a non-negative number at least L times.

Ideas

The data for this question is small, consider the direct and ordinary dp probability.

Then although $K$ can reach$2000$ , but can be regarded as$200\; to$ see. (Anyway, neither of them will explode the space and affect the result, so that it can also make$K$ becomes$200$ ) In other words, at any time$K$ value is greater than$At\; 200$ , you can change it to$200$ . (Of course, negative is to$-200$ )
Then we can easily set$f_{i,j,k}$为前iijjsucceeded in$i$ tasks$j$ , now$K$ value is$k$ 的概率。
那得到转移也很容易想到：
$\{\begin{array}{cc}{f}_{i,j,k+a_i}=f_{i-1,j,k+a_i}\times (1.0-p_i)& (j=0)\\ f_{i,j,k+a_i}=f_{i-1,j,k+a_i}\times (1.0-p_i)+f_{i-1,j-1,k}\times p_i& (j>0)\end{array}$
Of course, remember $The\; k$ value should be equal to$200$ takes the minimum value.
And you will findkkThe range of$k$ is$-200\sim 200$ , then all$k$ plus$200$ (all become non-negative before they can be placed in the array), and when the minimum value is reached, it is followed by$400$ ratio.

As for initialization, according to the above, $k$ plus$200$ , it should be$f_{0,0,K}=1$ becomes$f_{0,0,K+200}=1$ .
(In the beginning$k$ is$K$ is because ofKK in thebeginning$K$ is this value)

Then after dp is over, let's see how to count the answers.
It is easy to think that we enumerate the number of successful tasks in the last $i$ , the range is$L\sim n$ , thenKKat the end of the enumerationThe value of$K$$j$ , the range is$0\sim 200(+200)$ , then all$f_{n,i,j}$ Adding is the answer we want.

Code

#include<cstdio>
#include<iostream>
#define bp_add 200

using namespace std;

int n, l, K, a[201];
double p[201], f[201][201][402], ans;

int main() {
    
    
	scanf("%d %d %d", &n, &l, &K);
	
	for (int i = 1; i <= n; i++) {
    
    
		scanf("%lf", &p[i]);
		p[i] /= 100.0;
	}
	
	for (int i = 1; i <= n; i++) {
    
    
		scanf("%d", &a[i]);
	}
	
	f[0][0][K + bp_add] = 1.0;
	for (int i = 1; i <= n; i++)
		for (int j = 0; j <= i; j++)
			for (int k = -200; k <= 200; k++) {
    
    
				f[i][j][min(400, k + bp_add + a[i])] = f[i - 1][j][min(400, k + bp_add + a[i])] * (1.0 - p[i]);
				if (j) {
    
    
					f[i][j][min(400, k + bp_add + a[i])] += f[i - 1][j - 1][min(400, k + bp_add)] * p[i];
				}
			}
	
	for (int i = l; i <= n; i++)
		for (int j = bp_add + 0; j <= 200 + bp_add; j++)
			ans += f[n][i][j];
	
	printf("%.6lf", ans);
	
	return 0;
}