题目描述:
Goldbach’s conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:
Every even integer, greater than 2, can be expressed as the sum of two primes [1].
Now your task is to check whether this conjecture holds for integers up to 10^7.
输入:
Input starts with an integer T (≤ 300), denoting the number of test cases.
Each case starts with a line containing an integer n (4 ≤ n ≤ 10^7, n is even).
输出:
For each case, print the case number and the number of ways you can express n as sum of two primes. To be more specific, we want to find the number of (a, b) where
1 Both a and b are prime
2 a + b = n
3 a ≤ b
样例输入:
2
6
4
样例输出:
Case 1: 1
Case 2: 1
题目大意:
给出几组测试数据,每组给出一个n,问n能被分成几对素数的和。
思路:先进行一个素数打表,把数据范围内所有素数存在一个数字内,当然此时已经是从小到大排好的了,然后从数组中的第一个1到最后一个遍历,如果n减去该元素的值还是一个素数的话num++,如果该元素大于等于n/2+1,结束遍历。输出num的值即可。
code:
#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxn=1e7+5;
bool visit[maxn];
int prime[1000000];
int cnt=0;
int main()
{
for(int i=2;i<=10000000; i++)
{
if(visit[i]==false)
{
prime[cnt++]=i;
for(int j=i+i; j<=10000000; j=j+i)
visit[j]=true;
}
}
visit[0]=visit[1]=true;
int tt;
scanf("%d",&tt);
for(int kk=1;kk<=tt;kk++)
{
int n;
scanf("%d",&n);
int sum=0;
for(int i=0;i<cnt;i++)
{
if(prime[i]>=n/2+1)break;
if(visit[n-prime[i]]==false)sum++;
}//直接使用visit会超时
printf("Case %d: %d\n",kk,sum);
}
return 0;
}