vector 可变长数组 hdu 3823

Problem F

Time Limit : 4000/2000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 6 Accepted Submission(s) : 1

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Problem Description

Besides the ordinary Boy Friend and Girl Friend, here we define a more academic kind of friend: Prime Friend. We call a nonnegative integer A is the integer B’s Prime Friend when the sum of A and B is a prime.
So an integer has many prime friends, for example, 1 has infinite prime friends: 1, 2, 4, 6, 10 and so on. This problem is very simple, given two integers A and B, find the minimum common prime friend which will make them not only become primes but also prime neighbor. We say C and D is prime neighbor only when both of them are primes and integer(s) between them is/are not.

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case only contains two integers A and B.

Technical Specification

1. 1 <= T <= 1000
2. 1 <= A, B <= 150

Output

For each test case, output the case number first, then the minimum common prime friend of A and B, if not such number exists, output -1.

Sample Input

22 43 6

Sample Output

Case 1: 1Case 2: -1


题意：

       

         即求一个数，能使a，b与之相加后，成为素数，并且a与b之间没有其他的素数。

 

 做法：

   

         由于无论a，b与什么数相加之间的差值都相等，所以实际上求的是大于b的与其前一个数的差值是a-b的素数。

 

        所以该题的关键是将20000000之前的素数打表，然后求其每个之间的差值，相等的存放到同一个数组中。

 

  这里主要介绍的是vertor 可变长数组的用法：

 

         初始化：vertor<int> a[100];               //即开的是二维数组，可以包含100行，每行的长度可以不同。

 

                     a[2].push_back(i);                    //将i放入第二行的尾部

 

                     a[2].size();                           //获得第二行的长度

 

 

代码实现：

 

#include<iostream>#include<vector>#include<stdio.h>#include<string.h>using namespace std;int a[19000000];bool b[20000001];vector <int>sub[155];int main(){    memset(b,0,sizeof(b));    b[1]=1;    for(int i=2;i*i<=20000000;i++){            if(b[i]==1)                  continue;            for(int j=i*2;j<=20000000;j+=i){                    b[j]=1;                    }                    }    int times=0;    for(int k=2;k<=20000000;k++){            if(b[k]==0){                a[times]=k;                //cout<<k<<endl;                times++;                }                }    for(int g=0;g<times-1;g++){            int temp=a[g+1]-a[g];            if(temp>150)                continue;            sub[temp].push_back(a[g+1]);            }    int m;    cin>>m;    int sum=0;    while(m--){       sum++;       long long x,y;       cin>>x>>y;       int temp1=abs(x-y);       int temp2=x>y?x:y;       long long res=0;       for(int d=0;d<sub[temp1].size();d++){               if(sub[temp1][d]>=temp2){                      res=sub[temp1][d];                      break;                      }                      }       cout<<"Case "<<sum<<": ";       if(res==0)           cout<<"-1"<<endl;       else                  cout<<res-temp2<<endl;           }    //system("pause");    return 0;}