本文主要是介绍sdut2623--The number of steps（概率dp第一弹，求期望），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
The number of steps
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描述
Mary stands in a strange maze, the maze looks like a triangle（the first layer have one room，the second layer have two rooms，the third layer have three rooms …）. Now she stands at the top point(the first layer), and the KEY of this maze is in the lowest layer’s leftmost room. Known that each room can only access to its left room and lower left and lower right rooms .If a room doesn’t have its left room, the probability of going to the lower left room and lower right room are a and b (a + b = 1 ). If a room only has it’s left room, the probability of going to the room is 1. If a room has its lower left, lower right rooms and its left room, the probability of going to each room are c, d, e (c + d + e = 1). Now , Mary wants to know how many steps she needs to reach the KEY. Dear friend, can you tell Mary the expected number of steps required to reach the KEY?
输入
输出
示例输入
3 0.3 0.7 0.1 0.3 0.6 0
示例输出
3.41
提示
来源
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
double dp[100][100] ;
double a , b , c , d , e ;
int i , j , n ;
int ff(int x,int y)
{
if( x <= n && y >=(n+1)-x )
return 1 ;
return 0 ;
}
void f()
{
return ;
}
int main()
{
while(scanf("%d", &n) && n)
{
scanf("%lf %lf", &a, &b);
scanf("%lf %lf %lf", &c, &d, &e);
memset(dp,0,sizeof(dp));
for(i = n ; i >= 1 ; i--)
{
for(j = (n+1)-i ; j <= n ; j++)
{
if(i == n && j == (n+1)-i) continue ;
else if( i == n )
dp[i][j] = 1.0*( dp[i][j-1] ) + 1.0 ;
else
{
if( j == (n+1)-i )
dp[i][j] = a*dp[i+1][j-1] + b*dp[i+1][j] + 1.0 ;
else
dp[i][j] = c*dp[i+1][j-1] + d*dp[i+1][j] + e*dp[i][j-1] + 1.0 ;
}
}
}
printf("%.2lf\n", dp[1][n]);
}
return 0;
}
这篇关于sdut2623--The number of steps（概率dp第一弹，求期望）的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
