• 201805牛客模拟考


    第一题AC,注意点:输入是long类型,不能用BufferedReader,不然很麻烦:

     1 public class Main {
     2 
     3     public static void main(String[] args) throws IOException {
     4         Scanner in = new Scanner(System.in);
     5         long x = in.nextLong();
     6         long f = in.nextLong();
     7         long d = in.nextLong();
     8         long p = in.nextLong();
     9         long last = d - x * f;
    10         if(last > 0) {
    11             long day = last / (x + p);
    12             System.out.println(day + f);
    13         }
    14         //如果钱还不够付初始房费的
    15         else {
    16             long day = d / x;
    17             System.out.println(day);
    18         }
    19     }
    20 
    21 }
    View Code

    第二题过了60%,穷举所有可能性。先计算单独吃一盒三份、一盒两份和一盒一份可维持的天数,然后再将剩余的数量拼凑起来。不知道哪里错了。代码如下:

     1 public class Main {
     2     public static void main(String[] args) {
     3         Scanner in = new Scanner(System.in);
     4         int t = in.nextInt();
     5         while(t-- != 0) {
     6             int n = in.nextInt();
     7             int a = in.nextInt();
     8             int b = in.nextInt();
     9             int c = in.nextInt();
    10             int day_c = c / 2, last_c = c % 2;
    11             int day_b = b / 3, last_b = b % 3;
    12             int day_a = a / 6, last_a = a % 6;
    13             int last = 0;
    14             //三份还剩一盒
    15             if(last_c == 1) {
    16                 //两份还剩一盒
    17                 if(last_b == 1) {
    18                     //3+2+1
    19                     if(last_a > 0) {
    20                         last = 1;
    21                     }
    22                 }
    23                 //两份还剩两盒
    24                 else if(last_b == 2) {
    25                     //3+2+1
    26                     if(last_a > 0 && last_a < 5) {
    27                         last = 1;
    28                     }
    29                     //3+4+5
    30                     else if(last_a == 5) {
    31                         last = 2;
    32                     }
    33                 }
    34                 else {
    35                     //3+0+3
    36                     if(last_a >= 3) {
    37                         last = 1;
    38                     }
    39                 }
    40             }
    41             else {
    42                 //0+2+4
    43                 if(last_b == 1) {
    44                     if(last_a >= 4) {
    45                         last = 1;
    46                     }
    47                 }
    48                 //0+4+2
    49                 else if(last_b == 2) {
    50                     if(last_a >= 2) {
    51                         last = 1;
    52                     }
    53                 }
    54             }
    55         //    System.out.println(day_c + "," + day_b + "," + day_a + "," + last);
    56             int day = day_c + day_b + day_a + last;
    57             if(day >= n) {
    58                 System.out.println("Yes");
    59             }
    60             else {
    61                 System.out.println("No");
    62             }
    63         }
    64     }
    65 }
    View Code

    据说别人AC的思路,也是穷举,只是先单独吃一盒三份和一盒两份的,然后再将剩余的与一盒一份的进行拼凑,最后单独吃一盒一份的。代码如下:

     1 public class Main2 {
     2 
     3     public static void main(String[] args) {
     4         Scanner in = new Scanner(System.in);
     5         int t = in.nextInt();
     6         while(t-- != 0) {
     7             int n = in.nextInt();
     8             int a = in.nextInt();
     9             int b = in.nextInt();
    10             int c = in.nextInt();
    11             int day_c = c / 2, last_c = c % 2;
    12             int day_b = b / 3, last_b = b % 3;
    13             int day = 0;
    14             //三份还剩一盒
    15             if(last_c == 1) {
    16                 //两份还剩一盒
    17                 if(last_b == 1) {
    18                     //3+2+1
    19                     if(a >= 1) {
    20                         day++;
    21                         a -= 1;
    22                     }
    23                 }
    24                 //两份还剩两盒
    25                 else if(last_b == 2) {
    26                     //3+2+1
    27                     if(a >= 1 && a < 5) {
    28                         a -= 1;
    29                         day++;
    30                     }
    31                     //3+4+5
    32                     else if(a >= 5) {
    33                         a -= 5;
    34                         day += 2;
    35                     }
    36                 }
    37                 else {
    38                     //3+0+3
    39                     if(a >= 3) {
    40                         a -= 3;
    41                         day++;
    42                     }
    43                 }
    44             }
    45             else {
    46                 //0+2+4
    47                 if(last_b == 1) {
    48                     if(a >= 4) {
    49                         a -= 4;
    50                         day++;
    51                     }
    52                 }
    53                 //0+4+2
    54                 else if(last_b == 2) {
    55                     if(a >= 2) {
    56                         a -= 2;
    57                         day++;
    58                     }
    59                 }
    60             }
    61             day = day + day_c + day_b + a / 6;
    62             if(day >= n) {
    63                 System.out.println("Yes");
    64             }
    65             else {
    66                 System.out.println("No");
    67             }
    68         }
    69     }
    70 
    71 }
    View Code
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  • 原文地址:https://www.cnblogs.com/cing/p/9083628.html
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