Sorry About That, Chief!（阅读理解题）

题目描述

When Dr. Orooji was your age, one of the popular TV shows was “Get Smart!” The main character in this show (Maxwell Smart, a secret agent) had a few phrases; we used one such phrase for the title of this problem and we’ll use couple more in the output description!

A “prime” number is an integer greater than 1 with only two divisors: 1 and itself; examples include 5, 11 and 23. Given a positive integer, you are to print a message indicating whether the number is a prime or how close it is to a prime.

输入

The first input line contains a positive integer, n (n ≤ 100), indicating the number of values to check. The values are on the following n input lines, one per line. Each value will be an integer between 2 and 10,000 (inclusive).

输出

For each test case, output two integers on a line by itself separated by a space. The first integer is the input value and the second integer is as follows:

• If the input number is a prime, print 0 (zero). This refers to Maxwell Smart phrase: “Would you believe it; it is a prime!”
• If the input number is not a prime, print the integer d where d shows how close the number is to a prime number (note that the closest prime number may be smaller or larger than the given number). This refers to Maxwell Smart phrase: “Missed it by that much (d)!”

样例输入

4
23
25
22
10000

样例输出

23 0
25 2
22 1
10000 7

题意：
给出一个数n，然后下面是n个数
对于下面这n个数，对每个数进行判断，如果这个数是素数，就输出按照对应的格式输出这个数然后再输出0；相反，如果这个数不是素数，就输出这个数和离这个数最近的一个素数的距离；
本蒟蒻看完这个题的时候看到这个数据范围很想进行一波暴力，可是想了想，如果暴力代码会又臭又长，并且我的代码和别人的代码本来就又臭又长，从而进行筛一下！

#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize (2)
#pragma G++ optimize (2)
#include <bits/stdc++.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define wuyt main
typedef long long ll;
#define HEAP(...) priority_queue<__VA_ARGS__ >
#define heap(...) priority_queue<__VA_ARGS__,vector<__VA_ARGS__ >,greater<__VA_ARGS__ > >
template<class T> inline T min(T &x,const T &y){return x>y?y:x;}
template<class T> inline T max(T &x,const T &y){return x<y?y:x;}
//#define getchar()(p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 21) + 1], *p1 = buf, *p2 = buf;
ll read(){ll c = getchar(),Nig = 1,x = 0;while(!isdigit(c) && c!='-')c = getchar();
if(c == '-')Nig = -1,c = getchar();
while(isdigit(c))x = ((x<<1) + (x<<3)) + (c^'0'),c = getchar();
return Nig*x;}
#define read read()
const ll inf = 1e15;
const int maxn = 2e5 + 7;
const int mod = 1e9 + 7;
#define start int wuyt()
#define end return 0
/**bool prime(int n)
{
    if(n==1||n==1)
        return false;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0) return false;
    }
    return true;
}///想要暴力的动机体现的淋漓尽致**/
int num[maxn],n,cnt,judge[maxn];
void shai()
{
    for(int i=2;i<=maxn;i++)
    {
        if(!judge[i])
            num[++cnt]=i;
        for(int j=1;j<=cnt&&num[j]*i<=maxn;j++)
        {
            judge[i*num[j]]=1;
            if(i%num[j]==0)
                break;
        }
    }
}
start{
    int cnt1=read;
    shai();
    while(cnt1--)
    {
        int minn=mod;
        int number=read;
        if(judge[number]==0) printf("%d 0
",number);
        if(judge[number]!=0){
            for(int i=1;i<=10009;i++)
            {
                if(judge[i]==0)
                    minn=min(minn,abs(i-number));
            }
            printf("%d %d
",number,minn);
        }
    }
	end;
}

说明几点，这里的judge数组里面存放的是对应下标是不是素数；是素数为0，不是素数为1
另外呢对于这里：

for(int i=1;i<=10009;i++)
{
    if(judge[i]==0)
        minn=min(minn,abs(i-number));
}

这里的操作实在是有点粗鲁 ，完全可以进行优化一下！！！