浙工大姗姗杯round2 G - Bad Luck Island CodeForces - 540D
2017-07-16 13:35
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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.
Examples
input
output
input
output
input
output
题意:岛上生活着包、剪、锤三种生物,两两相遇遵照石头剪刀布法则杀死对方。现在给你岛上初始包剪锤各自的数量,求包剪锤各自最后的存活率。
思路:概率dp。本来以为有公式可以推到,但是想想不可能。dp[i][j][k],i,j,k分别表示包剪锤数量,i*j+i*k+j*k是总的相遇情况,又因为两种物种相遇
4000
死亡方已知,所以可以用i*j/(i*j+i*k+j*k)来表示概率。接下来就是dp
#include <stdio.h>
#include <iostream>
using namespace std;
double dp[128][128][128];
int main(){
int r, s, p;
cin >> r >> s >> p;
for(int i = 1; i <= 127; i++)
for(int j = 0; j <= 127; j++)
dp[i][j][0] = 1;
for(int i = 1; i <= 127; i++)
for(int j = 1; j <= 127; j++)
for(int k = 1; k <= 127; k++){
double p = i*j + j*k + i*k;//总情况数;
dp[i][j][k] += i*k*dp[i - 1][j][k];
dp[i][j][k] += i*j*dp[i][j - 1][k];
dp[i][j][k] += j*k*dp[i][j][k - 1];
dp[i][j][k] /= p;
}
printf("%.12lf %.12lf %.12lf\n", dp[r][s][p], dp[s][p][r], dp[p][r][s]);
return 0;
}
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.
Examples
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
题意:岛上生活着包、剪、锤三种生物,两两相遇遵照石头剪刀布法则杀死对方。现在给你岛上初始包剪锤各自的数量,求包剪锤各自最后的存活率。
思路:概率dp。本来以为有公式可以推到,但是想想不可能。dp[i][j][k],i,j,k分别表示包剪锤数量,i*j+i*k+j*k是总的相遇情况,又因为两种物种相遇
4000
死亡方已知,所以可以用i*j/(i*j+i*k+j*k)来表示概率。接下来就是dp
#include <stdio.h>
#include <iostream>
using namespace std;
double dp[128][128][128];
int main(){
int r, s, p;
cin >> r >> s >> p;
for(int i = 1; i <= 127; i++)
for(int j = 0; j <= 127; j++)
dp[i][j][0] = 1;
for(int i = 1; i <= 127; i++)
for(int j = 1; j <= 127; j++)
for(int k = 1; k <= 127; k++){
double p = i*j + j*k + i*k;//总情况数;
dp[i][j][k] += i*k*dp[i - 1][j][k];
dp[i][j][k] += i*j*dp[i][j - 1][k];
dp[i][j][k] += j*k*dp[i][j][k - 1];
dp[i][j][k] /= p;
}
printf("%.12lf %.12lf %.12lf\n", dp[r][s][p], dp[s][p][r], dp[p][r][s]);
return 0;
}
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