验证哥德巴赫猜想:一个不小于6 的偶数可以表示为两个素数之和,如6=3+3,8=3+5,10=3+7

验证哥德巴赫猜想:一个不小于6 的偶数可以表示为两个素数之和,如6=3+3,8=3+5,10=3+7

#include stdio.hbool prime(int n) {for(int i = 2; i if(n%i==0) return false;return true;}int main () {int m, n, count = 0;scanf(%d%d, &m, &n);for(int i = m; i for(int j = 2; j if (prime(j)&&(i-j)) { printf(%d=%d+%d , i, j, i-j); count++; if(count%5 == 0) printf(\n); } }}return 0;} #include stdio.h#include math.hint main(void){int count, i, j, k, m, n, number;scanf(%d%d, &m, &n);if(m % 2 != 0) m = m + 1;if(m >= 6){ count = 0; for(i = m; i for(j = 2; j number= 1; for(k=2;k if(j%k==0) number = 0; for(k=2;k if((i-j)%k==0) number = 0; if(number == 1) { printf(%d=%d+%d , i, j, i-j); count++; if(count%5 == 0) printf(\n); break; } } }}}居然是填空题,疯了……======以下答案可供参考======供参考答案1:表格（三）：偶数表为两个奇质数之和的表法的数量——公式计算数与实际表法数之比Gc（N）≡ Ctwin × K × 4 / N × Sha（N/2）×（ Sha（N）－ Sha（N/2））; K ≡ 4.7593318809Ctwin ≡ 0.6601618158468; Sha（N）≡ 2 /（1＋√（1－4/Ln（N）））× N/ Ln（N）HNGp（N）Sha（N）Sha（N/2）Gc（N）Gc / Gp1646969323043755729300333274.349155171875.98443755922.8541.0000044303821293938646082116003581899754.297300333274.34982134780.5051.00022867047319409079690118808995857185414.630442158083.119118824420.8391.000129837304258787729201544612911128545184.65581899754.297154477985.9621.000108085085323484661501893600431397058598.85720132249.812189389705.7791.000156647516388181593802237100821663346502.21857185414.630223726819.0721.000074815907452878526102575557311927811728.25993273148.864257596871.3481.000159733778517575458402910421932190736390.321128545184.65291074634.3351.000111466099582272390703241858642452328817.441263112051.68324214987.3171.00008983524

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