Zoj 3329 one person game (probability of loops DP)

Http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 3754

I started to read the wrong question and thought that I didn't go to A, B, and C. I had to add A + B + C on the basis of the original one. Then I came up with a formula, I did not expect that the sample has passed, so I could not debug it.

The formula is very good. If DP [I] is set, it indicates the expectation to be throttled when the current time is I, and can be transferred to two States: DP [0] and DP [I + K]. Transfer to DP [0]

The probability is P0, and the probability of transferring to DP [I + k] is PK. DP [I] = P0 * DP [0] + PK * DP [I + k] + 1. It is found that all DP [I] are related to DP [0.

The following derivation references: http://www.cnblogs.com/kuangbin/archive/2012/10/03/2710648.html

Set DP [I] = A [I] * DP [0] + B [I]. The following result is obtained from the right side of the above equation: DP [I] = Σ (Pk * A [I + k] * DP [0] + PK * B [I + k]) + dp [0] * P0 + 1 = (Σ (Pk * A [I + k]) + P0) DP [0] + Σ (Pk * B [I + k]) + 1; obvious a [I] = (Σ (Pk * A [I + k]) + P0) B [I] = Σ (Pk * B [I + k]) + 1 first recursion to obtain a [0] and B [0]. then DP [0] = B [0]/(1-A [0]);

Then, a [0] and B [0] are reversed.

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>//#define LL __int64#define LL long long#define eps 1e-8#define PI acos(-1.0)using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 4010;double dp[10010];double A[600],B[600];double p[100];int main(){int test;int n,a,b,c;int k1,k2,k3;scanf("%d",&test);while(test--){scanf("%d %d %d %d %d %d %d",&n,&k1,&k2,&k3,&a,&b,&c);double p0 = 1.0/(k1*k2*k3);memset(p,0,sizeof(p));for(int i = 1; i <= k1; i++){    for(int j = 1; j <= k2; j++)    {        for(int k = 1; k <= k3; k++)        {            if(i != a || j != b || k != c)                        p[i+j+k] += p0;        }    }}memset(A,0,sizeof(A));memset(B,0,sizeof(B));for(int i = n; i >= 0; i--){    A[i] = p0,B[i] = 1;    for(int j = 1; j <= k1+k2+k3; j++)    {        A[i] += p[j]*A[i+j];        B[i] += p[j]*B[i+j];    }}printf("%.15lf\n",B[0]/(1-A[0]));}return 0;}

Zoj 3329 one person game (probability of loops DP)