(一)算法之暴力破解法

xiaoxiao2021-02-28  115

1.暴力破解 public class Baolipojie {      /**       * 鸡兔同笼       * 设鸡为x 兔为y       */      @Test      public void test1() {            int HEAD = 50;            int FOOT = 120;            for (int x = 0; x < 50; x++) {                 int y = HEAD - x;                 if(x+y==HEAD && 2*x+4*y==FOOT) {                      System.out.println("鸡的数量:"+x+"兔的数量为:"+y);                 }            }      }      /**       * 韩信点兵       * 韩信部队人数大概1000多人,5人一组还剩1人,7人一组还剩2人,8人一组还剩3人       *       */      @Test      public void test2() {            for (int i = 1000; i < 2000; i++) {                 if(i%5==1 && i%7==2 && i%8==3) {                      System.out.println("韩信部队人数大概为:"+i);                 }            }      }
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