大家好，又见面了，我是全栈君。
The number of steps
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描写叙述
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
演示样例输入
3 0.3 0.7 0.1 0.3 0.6 0
演示样例输出
3.41
提示
来源
ij,
i到i的概率是p
ii
,期望是E，那么求1到4的期望是
#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;
}
发布者：全栈程序员-用户IM，转载请注明出处：https://javaforall.cn/115755.html原文链接：https://javaforall.cn
【正版授权，激活自己账号】： Jetbrains全家桶Ide使用，1年售后保障，每天仅需1毛
【官方授权 正版激活】： 官方授权 正版激活 支持Jetbrains家族下所有IDE 使用个人JB账号...