HDOJ 1493 QQpet exploratory park

QQpet exploratory park

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 465    Accepted Submission(s): 244

Problem Description

Today, more and more people begin to raise a QQpet. You can get a lot of pleasure from it, although it does not have a real life and it calls for huge patience to take care of it. There is a place called QQpet exploratory park in the world. Every week, you can get a chance to have a joy there for free. The whole park contains 61 grids in a line, numbered from 0 to 60. Ten of them are important grids which will touch off ( 引发 ) an incident when the pet stands on. They are 5, 12, 22, 29, 33, 38, 42, 46, 50 and 55. Your pet is standing on the gird of number 0 in the beginning. You can toss the die ( 掷骰子 ) 10 times. Each time, the pet goes ahead n steps which n is the number from the die ( n ∈{ 1, 2, …, 6 } ). If your RP is great enough( calls RPG for short ), you will get many surprises in the important grids, such as some yuanbao( the money in QQpet world ), an improvement of your pet's ability, and the most attractive gift-package. Now, your task is to calculate the probability(概率) of touching each important grid.

 

Input

The first line of the input contains an integer t– determining the number of datasets. Then t lines follows. Each line contains 6 numbers pi, i ∈{ 1, 2, …, 6 }, indicating the probability of getting 1 to 6 after you toss the die every time . p1+ p2+ … + p6 = 1.

 

Output

For each test case, output the probability of touching each important grid. accurate up to 1 decimal places. There is a blank line between test cases. See the Sample Output to get the exactly output format.

 

Sample Input

20.000 1.000 0.000 0.000 0.000 0.000
0.500 0.000 0.000 0.000 0.000 0.500

 

Sample Output

5: 0.0%
12: 100.0%
22: 0.0%
29: 0.0%
33: 0.0%
38: 0.0%
42: 0.0%
46: 0.0%
50: 0.0%
55: 0.0%

5: 3.1%
12: 30.5%
22: 27.3%
29: 24.6%
33: 21.9%
38: 10.9%
42: 0.8%
46: 0.0%
50: 4.4%
55: 1.0%

 

Author

LL

 

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

 

Recommend

LL

 

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

double p[7];

double f[11][66];

void dp()

{

    for(int k=1;k<=10;k++)

    {

        for(int j=60;j>=1;j--)

        {

            for(int i=1;i<=6;i++)

            {

                if(j>=i)

                f[k][j]+=f[k-1][j-i]*p ;

            }

        }

    }

}

int main()

{

    int T;

    cin>>T;

    while(T--)

    {

        memset(f,0,sizeof(f));

        f[0][0]=1.0;

        for(int i=1;i<=6;i++)

            scanf("%lf",&p );

        dp();

        double ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [5];

        printf("5: %.1lf%%\n",ans*100);

        ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [12];

        printf("12: %.1lf%%\n",ans*100);

        ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [22];

        printf("22: %.1lf%%\n",ans*100);

        ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [29];

        printf("29: %.1lf%%\n",ans*100);

        ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [33];

        printf("33: %.1lf%%\n",ans*100);

        ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [38];

        printf("38: %.1lf%%\n",ans*100);

                ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [42];

        printf("42: %.1lf%%\n",ans*100);

                ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [46];

        printf("46: %.1lf%%\n",ans*100);

                ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [50];

        printf("50: %.1lf%%\n",ans*100);

                ans=0;

        for(int i=1;i<=10;i++)

            ans+=f [55];

        printf("55: %.1lf%%\n",ans*100);

        if(T)

            putchar(10);

    }

    return 0;

}

转载于:https://www.cnblogs.com/CKboss/archive/2013/06/11/3351023.html