Goldbach`s Conjecture
2014-05-29 21:25
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Description
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 107.
Input
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 ≤ 107, n is even).
Output
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
Sample Input
2
6
4
Sample Output
Case 1: 1
Case 2: 1
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 107.
Input
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 ≤ 107, n is even).
Output
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
Sample Input
2
6
4
Sample Output
Case 1: 1
Case 2: 1
#include<stdio.h> #include<string.h> bool b[10000005]={0};//不能是int int a[1000005]={0}; void init() { int i,j; for(i=2;i<=10000003;i++) { if(!b[i]) for(j=i+i;j<=10000003;j+=i) b[j]=1; } j=1; for(i=2;i<=10000003;i++) if(b[i]==0) a[j++]=i; } int main() { int i,j,m,n,ca=1,t; init(); scanf("%d",&t); while(t--) { scanf("%d",&n); m=0; for(i=1;i<=1000002;i++) { if(a[i]>n/2)break; if(b[n-a[i]]==0)m++; } printf("Case %d: %d\n",ca++,m); } return 0; }
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