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NYOJ--517--最小公倍数(大数打表)

發布時間:2024/6/21 编程问答 36 豆豆
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最小公倍數

時間限制:1000?ms ?|? 內存限制:65535?KB 難度:3 描述
為什么1小時有60分鐘,而不是100分鐘呢?這是歷史上的習慣導致。 但也并非純粹的偶然:60是個優秀的數字,它的因子比較多。 事實上,它是1至6的每個數字的倍數。即1,2,3,4,5,6都是可以除盡60。 我們希望尋找到能除盡1至n的的每個數字的最小整數m.
輸入
多組測試數據(少于500組)。
每行只有一個數n(1<=n<=100).
輸出
輸出相應的m。
樣例輸入
2 3 4
樣例輸出
2 6 12
1 //打表 2 import java.math.BigDecimal; 3 import java.math.BigInteger; 4 import java.util.Scanner; 5 public class Main{ 6 7 public static void main(String args[]){ 8 Scanner cin = new Scanner(System.in); 9 /*final int MAX = 105; 10 int arr[] = new int[MAX]; 11 BigInteger res[] = new BigInteger[MAX]; 12 for(int i=1; i<MAX; ++i)arr[i] = i; 13 for(int i=2; i<MAX; ++i){ 14 for(int j=i+1; j<MAX; ++j){ 15 if(j%i == 0) 16 arr[j] /= arr[i]; 17 } 18 } 19 for(int i=1; i<MAX; ++i)res[i] = BigInteger.ONE; 20 for(int i=2; i<MAX; ++i){ 21 for(int j=2; j<i; ++j){ 22 res[i] = res[i].multiply(BigInteger.valueOf(arr[j])); 23 } 24 } 25 for(int i=1; i<101; ++i){ 26 int n = i; 27 System.out.println("\""+res[n+1] + "\","); 28 }*/ 29 String s[] = { 30 "1", 31 "2", 32 "6", 33 "12", 34 "60", 35 "60", 36 "420", 37 "840", 38 "2520", 39 "2520", 40 "27720", 41 "27720", 42 "360360", 43 "360360", 44 "360360", 45 "720720", 46 "12252240", 47 "12252240", 48 "232792560", 49 "232792560", 50 "232792560", 51 "232792560", 52 "5354228880", 53 "5354228880", 54 "26771144400", 55 "26771144400", 56 "80313433200", 57 "80313433200", 58 "2329089562800", 59 "2329089562800", 60 "72201776446800", 61 "144403552893600", 62 "144403552893600", 63 "144403552893600", 64 "144403552893600", 65 "144403552893600", 66 "5342931457063200", 67 "5342931457063200", 68 "5342931457063200", 69 "5342931457063200", 70 "219060189739591200", 71 "219060189739591200", 72 "9419588158802421600", 73 "9419588158802421600", 74 "9419588158802421600", 75 "9419588158802421600", 76 "442720643463713815200", 77 "442720643463713815200", 78 "3099044504245996706400", 79 "3099044504245996706400", 80 "3099044504245996706400", 81 "3099044504245996706400", 82 "164249358725037825439200", 83 "164249358725037825439200", 84 "164249358725037825439200", 85 "164249358725037825439200", 86 "164249358725037825439200", 87 "164249358725037825439200", 88 "9690712164777231700912800", 89 "9690712164777231700912800", 90 "591133442051411133755680800", 91 "591133442051411133755680800", 92 "591133442051411133755680800", 93 "1182266884102822267511361600", 94 "1182266884102822267511361600", 95 "1182266884102822267511361600", 96 "79211881234889091923261227200", 97 "79211881234889091923261227200", 98 "79211881234889091923261227200", 99 "79211881234889091923261227200", 100 "5624043567677125526551547131200", 101 "5624043567677125526551547131200", 102 "410555180440430163438262940577600", 103 "410555180440430163438262940577600", 104 "410555180440430163438262940577600", 105 "410555180440430163438262940577600", 106 "410555180440430163438262940577600", 107 "410555180440430163438262940577600", 108 "32433859254793982911622772305630400", 109 "32433859254793982911622772305630400", 110 "97301577764381948734868316916891200", 111 "97301577764381948734868316916891200", 112 "8076030954443701744994070304101969600", 113 "8076030954443701744994070304101969600", 114 "8076030954443701744994070304101969600", 115 "8076030954443701744994070304101969600", 116 "8076030954443701744994070304101969600", 117 "8076030954443701744994070304101969600", 118 "718766754945489455304472257065075294400", 119 "718766754945489455304472257065075294400", 120 "718766754945489455304472257065075294400", 121 "718766754945489455304472257065075294400", 122 "718766754945489455304472257065075294400", 123 "718766754945489455304472257065075294400", 124 "718766754945489455304472257065075294400", 125 "718766754945489455304472257065075294400", 126 "69720375229712477164533808935312303556800", 127 "69720375229712477164533808935312303556800", 128 "69720375229712477164533808935312303556800", 129 "69720375229712477164533808935312303556800", 130 }; 131 while(cin.hasNext()){ 132 int n = cin.nextInt(); 133 System.out.println(s[n-1]); 134 } 135 } 136 }

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轉載于:https://www.cnblogs.com/evidd/p/6775167.html

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