2014 NEUACM 新星赛

1481: Sister Wang

时间限制: 1 Sec  内存限制: 256 MB
提交: 514  解决: 115
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题目描述

Do you know Sister Wang?

Sister Wang is goddess and many diaosi want to know her phone number. Sister Wang doesn’t tell them her number directly. She tells them a string, contains letters,

 digits and symbols,but no blanks. And her phone number is in the string. Fushuai knows how to get the phone number. He delete the letters and symbols from the 

string, so that he gets some digits, and then he sort the digits in ascending order. In this way, Fushuai succeeds in calling Sister Wang.

输入

There are several cases. Each cases contains a string L(1<= length(L)<=20).

输出

Output Sister Wang’s phont number.

样例输入

Sist1e2r3w4a5ng

样例输出

12345

提示

来源

新星赛2014

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AC代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int main(){char s[111];string str;while(~scanf("%s",s)){str="";for(int i=0;s[i]!='\0';i++){if(s[i]>='0'&&s[i]<='9')str=str+s[i];}sort(str.begin(),str.end());cout<<str<<endl;}return 0;}/**************************************************************    Problem: 1481    User: 1307122118    Language: C++    Result: 正确    Time:4 ms    Memory:1260 kb****************************************************************/

1482: Christmas math gift

时间限制: 1 Sec  内存限制: 256 MB
提交: 56  解决: 27
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题目描述

Wow,today is Christmas again!Jiajia is happy because he can go out to play with himself.So poor Jiajia because he can only play with himself while others has friend! Although Jiajia is so poor but his mother don’t let go out unless he finish her Christmas gift-----A math problem.So crazy,don’t you think so! Jiajia only look the math problem for a second and then he decide-------to give up because it’s so difficult.Mom give him a equation-----a*x^2+b*x+c=0 and ask him how many solution does the equation has. But Jiajia’s math skill didn’t improve after he graduated from the Kindergarten(Are you kidding me),so he throw the problem to you and rush out to play with himself(How happy Jiajia is ^_^),now your task is to solve the problem as soon as possible.

输入

The firs line is a integer t,the number of cases.(t<=10)

In each case,there are three integer,a,b,c (0<=a,b,c<=10^9)in a line,represent the coefficient of the equation.

输出

For each cases,you are required to output a line.

If there on solution,output “no solution”,if there is only one solution,output

“one solution”,if there are more than one solution, output “multiple solution”.

样例输入

23 3 30 0 0

样例输出

no solutionmultiple solution

提示

来源

新星赛2014

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AC代码：

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<algorithm>

#include<queue>

usingnamespace std;

intmain()

{

    intt,a,b,c;

    intdel;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d%d%d",&a,&b,&c);

        if(a==0)

        {

            if(b==0)

            {

                if(c==0)

                {

                    printf("multiple solution\n");

                }

                else

                {

                    printf("no solution\n");

                }

            }

            else

            {

                printf("one solution\n");

            }

        }

        else

        {

            del=b*b-4*a*c;

            if(del<0)

            {

                printf("no solution\n");

            }

            elseif(del==0)

            {

                printf("one solution\n");

            }

            else

            {

                printf("multiple solution\n");

            }

        }

    }

 

    return0;

}

/**************************************************************

    Problem: 1482

    User: 1307122118

    Language: C++

    Result: 正确

    Time:0 ms

    Memory:1252 kb

****************************************************************/

1483: Fushaui see movie

时间限制: 1 Sec  内存限制: 256 MB
提交: 83  解决: 25
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题目描述

It’s Christmas again! Li fushuai is handsome and wealthy, so many girl wants to see movie with him. It’s a difficult problem because Li fushuai every time can only see movie with one girl but he want to see movie with girl as many as possible. At the same time,Li fushuai very understand how to enjoy the life. Every girl has a charm value-----represent the happiness degree Li fushuai can get when he see movie with this girl. Now, the girls stand in a circle(What a spectacular scene!),and Li fushuai can’t move their position.Now Li fushuai will choose a queue of continuous girls in the circle as his company to see movie. Be care,in the queue, the girl stand in the front has the Priority to see movie with Li fushuai . Because Li fushuai know how to enjoy life so much that he require that the charm value of the girl stand at the back must greater strictly than the girl stand in the front. Li fushuai is lazy so he asks to help him find the queue satisfy the requirements above as long as possible.

输入

The first line is a integer t,imply the number of cases.(t<=10)

For each case,the first line is a integer n,imply the number of girls(n<=10^5),and the second lines have n integer,and the i-th integer indicate the charm value of the i-th girl

输出

For each case, print a integer,indicate the longest length of the queue.

样例输入

251 2 2 4 452 3 4 5 1

样例输出

25

提示

来源

新星赛2014

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AC代码：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>#include<queue> using namespace std;int a[200005],f[200005];int main(){int t,i,n,ans;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);a[i+n]=a[i];}f[0]=1;ans=1;for(i=1;i<2*n;i++){if(a[i]>a[i-1]){f[i]=f[i-1]+1;}else{f[i]=1;}ans=max(ans,f[i]);}printf("%d\n",ans);}    return 0;}

1484: Hengheng eat noodles

时间限制: 1 Sec  内存限制: 256 MB
提交: 15  解决: 12
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题目描述

As we all know,Hengheng(big cow),very look forward to meizi and he often declare that he want to find a meizi.As Hengheng’s admirer, Jiajia decides to help Hengheng to achieve his dream in Christmas Day. In this day, Jiajia asks Hengheng to eat noodles with a pretty meizi(don’t ask why Jiajia didn’t chase this pretty meizi). When eating noodles, Hengheng is so excited that he wants the qq of the meizi. However, a unlucky thing happen.The pretty meizi says she won’t give Hengheng her qq unless Hengheng can solve her question. Hengheng promise without hesition.The pretty meizi picks up a noodle and say with sexy voice, “We all know a noodle has two ends,and now I has n noodles, so there are 2*n ends at all. Out of interest,I will link these ends randomly,and then I wonder the mathematical expectation of the circle I could get.” Although Hengheng is big cow, he can’t wait to get the qq so he can’t think as long as he see the pretty. So he ask Jiajia to solve the problem for him. Luckily,after 1000 ms,Jiajia give the right answer. And Hengheng got the qq successfully. Now, if you are Jiajia, could solve the problem within 1000 ms and help Hengheng get the qq.

输入

The first line is a integer t,indicate the number of cases.(t<=10)
For each case follow,there is a integer n,imply the amount of noodles.(n<=10^6)

输出

For each case, print a double E,represent the mathematical expectation.(E retain 6 decimal)

样例输入

3124

样例输出

1.0000001.3333331.676190

提示

来源

新星赛2014

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题意：求n条面条围成圆圈的数学期望。

解题思想：n条面条总共有2n个端点。可看做n-1条面条增加一条面条，对于增加的这条面条，我们随便选取一个端点，总共有2n-1个端点可以与之结合：

1.对于第一种情况，只能跟自己另一条边组合，故第一种情况圆圈数加一，这种概率为p1=1/(2n-1)

2.除了第一种情况就是第二种情况，圆圈数不变，这种概率为p2=(2n-2)/(2n-1)

数学规律递推公式推导(E'代表连加符号西格玛,S表示圆圈数，p表示对应的概率)：

exp(n-1)=E'SjPj

exp(n)=E'(Sj+1)PjP1+E'SjPjP2

           =(E'SjPj+E'Pj)P1+E'SjPjP2

           =(exp(n-1)+1)P1+exp(n-1)P2

           =(exp(n-1)+1)*1/(2n-1)+exp(n-1)*(2n-2)/(2n-1)

AC代码：

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#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<queue>

usingnamespace std;

doublee[1000001];

intmain()

{

    intt,i,n;

    e[1]=1.000000;

    for(i=2;i<=1000001;i++)

    {

        e[i]=(e[i-1]+1.0)/(2*i-1)+e[i-1]*(2*i-2)/(2*i-1);

    }

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        printf("%lf\n",e[n]);

    }

    return0;

}

/**************************************************************

    Problem: 1484

    User: 1307122118

    Language: C++

    Result: 正确

    Time:119 ms

    Memory:9064 kb

****************************************************************/




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#include<iostream>

#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

#include<queue>

usingnamespace std;

doublee[1000001];

intmain()

{

    intt,i,n;

    e[1]=1.000000;

    for(i=2;i<=1000001;i++)

    {

        e[i]=(e[i-1]+1.0)/(2*i-1)+e[i-1]*(2*i-2)/(2*i-1);

    }

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        printf("%lf\n",e[n]);

    }

    return0;

}

/**************************************************************

    Problem: 1484

    User: 1307122118

    Language: C++

    Result: 正确

    Time:119 ms

    Memory:9064 kb

****************************************************************/