Problem A: Eming
Time Limit: 1 Sec Memory Limit: 32 MBSubmit: 28 Solved: 24
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Description
Eming is a contest hold by WHUACM training team. The aim is to select new members of the team.
Usually, the first problem is a very simple problem such as “a+b problem”. But this time, Xioumu is tired of this kind of problem, he decide to solve “a and b problem”.
Give you the result of a + b and a^2 – b^2, please output the result of a and b.
Input
There are several test cases. For each case there are two double numbers indicating a+b and a^2-b^2. You may assume that a+b will not be zero.
Output
For each case please output the result of a and b in one line, rounded to 2 decimal places.
Sample Input
Sample Output
HINT
#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
float x,y;
while(scanf("%f",&x)!=EOF)
{
scanf("%f",&y);
float c=y/x;
printf("%.2f %.2f
",(x+c)*0.5,(x-c)*0.5);
}
return 0;
}
Problem B: Arithmetic Progression
Time Limit: 1 Sec Memory Limit: 32 MBSubmit: 54 Solved: 24
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Description
“In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, … is an arithmetic progression with common difference of 2.”
- Wikipedia
This is a quite simple problem, give you the sequence, you are supposed to find the length of the longest consecutive subsequence which is an arithmetic progression.
Input
There are several test cases. For each case there are two lines. The first line contains an integer number N (1 <= N <= 100000) indicating the number of the sequence. The following line describes the N integer numbers indicating the sequence, each number will fit in a 32bit signed integer.
Output
For each case, please output the answer in one line.
Sample Input
Sample Output
HINT
思路：求最大等差数列长度，想清楚就可以做了。#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;
#define MAX 100005
unsigned short num[MAX];
unsigned short tnum[MAX];
bool bnum[MAX];
int main()
{
int n,i;
while(scanf("%d",&n)!=EOF)
{
int m=-1,count=1,t=0,maxlong=0;
for(i=1;i<=n;i++)
scanf("%d",&num[i]);
memset(tnum,0,sizeof(tnum));
memset(bnum,0,sizeof(bnum));
for(i=1;i<n;i++)
{
int temp=num[i+1]-num[i];
//cout<<"temp="<<temp<<endl;
if(temp==m)
{
count++;
//cout<<"count="<<count<<endl;
}
else
{
if(tnum[m]<count&&m!=-1)
{
bool f=true;
for(int j=1;j<=MAX;j++)
if(tnum[j]==count)
f=false;
tnum[m]=count;
if(f)
{
//cout<<m<<endl;
bnum[m]=true;
}
}
m=temp;
count=1;
}
}
for(i=1;i<=MAX;i++)
{
//printf("%d ",tnum[i]);
if(maxlong<tnum[i])
maxlong=i;
if(maxlong==tnum[i])
if(bnum[i])
maxlong=i;
}
//printf("
");
printf("%d
",maxlong);
}
return 0;
}
Problem E: Function
Time Limit: 1 Sec Memory Limit: 32 MBSubmit: 52 Solved: 26
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Description
Define a function f(n)=(f(n-1)+1)/f(n-2). You already got f(1) and f(2). Now, give you a number m, please find the value of f(m).
Input
There are several test cases. Each case contains three integers indicating f(1), f(2) and m ( 1 <= f(1), f(2), m <= 1000,000,000).
Output
For each case, please output the value of f(m), rounded to 6 decimal places.
Sample Input
Sample Output
HINT
思路：想啊想，看啊看，哇塞，数据怎么这么大o(╯□╰)o，1000000000，用数组都保存不了，怎么办？#include<iostream>
#include<stdio.h>
using namespace std;
double x=1,y=1;
double function(int m)
{
if(m==1)
return x;
if(m==2)
return y;
else
return (function(m-1)+1)/function(m-2);
}
int main()
{
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int m;
while(scanf("%lf",&x)!=EOF)
{
scanf("%lf%d",&y,&m);
if(m%5==0)
printf("%.6lf
",function(5));
else
printf("%.6lf
",function(m%5));
}
return 0;
}
PS:暑假第一场训练，水题就这么多吧。