Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem F: The primary problem.

Larry now lost most of his knowledge after spending a few years on deserted islands all over the place. When Ryan brought up the fact that every even number greater than 3 can represented as the sum of two prime numbers, he couldn't believe it! So now Larry is trying to come up with some kind of counterexample, and you can help him!

Input

Each line will contain an integer N greater than 3. The input will terminate when N is 0. N will not be bigger than 1 million.

Output

For each test case, output one way of summing two prime numbers. If there are more than one set of sums for which this is true, choose the set of sum the maximizes the difference between them. See the sample output. If a number cannot be represented as the sum of two prime number, print "NO WAY!" as below:

Sample Input

4
5
6
7
9
10
11
0

Sample Output

4:
2+2
5:
2+3
6:
3+3
7:
2+5
9:
2+7
10:
3+7
11:
NO WAY!

Note:
10 can be 3+7 or 5+5, and since 7-3 is bigger than 5-5, we choose 3+7.

---

#include <stdio.h>
#define maxL (1000000>>5)+1
int p[80000], pt = 0;
int mark[maxL];
int GET(int x) {
    return mark[x>>5]>>(x&31)&1;
}
void SET(int x) {
    mark[x>>5] |= 1<<(x&31);
}
void sieve() {
    int i, j, k;
    int n = 1000000;
    for(i = 2; i <= n; i++) {
        if(!GET(i)) {
            for(k = n/i, j = i*k; k >= i; k--, j -= i)
                SET(j);
            p[pt++] = i;
        }
    }
}
int main() {
    sieve();
    int n, i;
    while(scanf("%d", &n) == 1 && n) {
        printf("%d:\n", n);
        int hn = n/2, flag = 0;
        for(i = 0; i < pt; i++) {
            if(p[i] > hn)
                break;
            if(!GET(n-p[i])) {
                printf("%d+%d\n", p[i], n-p[i]);
                flag = 1;
                break;
            }
        }
        if(!flag)
            puts("NO WAY!");
    }
    return 0;
}