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64 lines
1.7 KiB
C
64 lines
1.7 KiB
C
/**
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* 1007. 素数对猜想
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* 让我们定义 dn 为：dn = pn+1 - pn，其中 pi 是第i个素数。显然有 d1=1 且对于n>1有
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* dn 是偶数。“素数对猜想”认为“存在无穷多对相邻且差为2的素数”。
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*
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* 现给定任意正整数N (< 10^5)，请计算不超过N的满足猜想的素数对的个数。
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*
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* 输入格式：每个测试输入包含1个测试用例，给出正整数N。
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*
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* 输出格式：每个测试用例的输出占一行，不超过N的满足猜想的素数对的个数。
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*
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* 输入样例：
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* 20
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* 输出样例：
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* 4
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**/
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#include <stdio.h>
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#include <math.h>
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int main()
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{
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int N;
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scanf("%d", &N);
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int paircount = 0;
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/* record three successive numbers if they are prime numbers, start
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from 2, 3, 4 */
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int iPrimeMinus2 = 1, iPrimeMinus1 = 1, iPrime;
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/* record the prime numbers before 350, which is lt sqrt(100000) */
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int primes[100] = {2, 3};
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int primecount = 2;
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for (int i = 4; i <= N; i++)
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{
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iPrime = 1;
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/* test if i is a prime number */
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int sqrt_i = (int)sqrt(i);
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for(int j = 0; primes[j] <= sqrt_i && j < primecount; j++)
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{
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/* only need to test on prime numbers before sqrt(i) */
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if(i % primes[j] == 0)
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{
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iPrime = 0;
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}
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}
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if(iPrime == 1)
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{
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if(primecount < 100)
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{
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primes[primecount] = i;
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}
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primecount++;
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if(iPrimeMinus2 == 1) /* a prime pair */
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{
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paircount++;
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}
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}
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/* shift the flags */
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iPrimeMinus2 = iPrimeMinus1;
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iPrimeMinus1 = iPrime;
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}
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printf("%d", paircount);
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return 0;
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}
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