CF# 301D Bad Luck Island(확률 dp+ 기억화)

3373 단어 DP동적 기획확률
D. Bad Luck Island
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
The Bad Luck Island is inhabited by three kinds of species: r rocks, s scissors and p papers. At some moments of time two random individuals meet (all pairs of individuals can meet equiprobably), and if they belong to different species, then one individual kills the other one: a rock kills scissors, scissors kill paper, and paper kills a rock. Your task is to determine for each species what is the probability that this species will be the only one to inhabit this island after a long enough period of time.
Input
The single line contains three integers r, s and p (1 ≤ r, s, p ≤ 100) — the original number of individuals in the species of rock, scissors and paper, respectively.
Output
Print three space-separated real numbers: the probabilities, at which the rocks, the scissors and the paper will be the only surviving species, respectively. The answer will be considered correct if the relative or absolute error of each number doesn't exceed 10 - 9.
Sample test(s)
input
2 2 2

output
0.333333333333 0.333333333333 0.333333333333

input
2 1 2

output
0.150000000000 0.300000000000 0.550000000000

input
1 1 3

output
0.057142857143 0.657142857143 0.285714285714

제목: r는 s, s는 p, p는 r를 먹고 세 개의 수량을 제시하며 마지막 세 가지가 각각 살아남을 확률을 구한다.
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i = a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi
") using namespace std; #define INF 0x3f3f3f3f #define N 105 int r,s,p; double dp[N][N][N]; double dfs(int pre,int now,int nest) //pre now,now nest,nest pre, pre { if(dp[pre][now][nest]>=0) return dp[pre][now][nest]; if(pre==0) return 0; if(nest==0) return pre ? 1:0; double ans=0; if(pre&&now) ans+=(pre*now*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre,now-1,nest); //pre now if(now&&nest) ans+=(now*nest*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre,now,nest-1); //now nest, if(pre&&nest) ans+=(pre*nest*1.0)/(pre*now+now*nest+pre*nest)*dfs(pre-1,now,nest); //nest pre dp[pre][now][nest]=ans; return ans; } int main() { int i,j,n; mem(dp,-1); while(~scanf("%d%d%d",&r,&s,&p)) { printf("%.10lf
",dfs(r,s,p)); printf("%.10lf
",dfs(s,p,r)); printf("%.10lf
",dfs(p,r,s)); } return 0; }

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