3 solutions

  • 1
    @ 2021-10-13 21:34:12

    1030 压力测试赛B题题解

    题目回顾

    Krik晚上失眠，躺着很无聊，于是突发奇想道阳台上去看星星。不失所望，那夜繁星当空、星河奔流。Kirk越看越精神，在繁星中发现了一个又一个的数字，刚学编程不久的Kirk想到，电脑上也有星星，何不操作一波呢！ 经过一番痛苦的敲键盘事件之后，终于Kirk电脑屏幕上出现了下图十个数字的星星表达式： t

    （本来想给文本的，但是文本太丑了） 为了让大家体会一下Kirk痛苦的敲代码事件，所以Kirk给你两个整数A和B，你的任务是计算A+B的结果并且使用星星表达式输出

    题目分析

    从题中所给出的信息，我们可以得出：

    1. 两个数字相加会得到一个结果，并把结果转化为特殊表达形式输出。
    2. 特殊表达式如图所示，由于CMD的特性，只能一行输出完再输出下一行，所以我们要对每个数做一个拆解，把每一行特定输出。
    3. 特别要注意的是，测试样例中给出，两个数字之间有一个空格，第二个数字后无空格。

    可行代码

    这里给大家贴三份代码 (注：不建议直接CV过去解决问题，打表考的是细心，自己实操一下是最好的 /笔芯) Kirk的垃圾代码（强烈建议看看后面两篇比较好的写法，同时这也是几种不同思路，不说了Kirk思路最差）

    #include <stdio.h>

void cout0(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("* *");
    else if (line == 3)
        printf("* *");
    else if (line == 4)
        printf("* *");
    else if (line == 5)
        printf("***");
}

void cout1(int line) {
    if (line == 1)
        printf(" * ");
    else if (line == 2)
        printf(" * ");
    else if (line == 3)
        printf(" * ");
    else if (line == 4)
        printf(" * ");
    else if (line == 5)
        printf(" * ");
}

void cout2(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("  *");
    else if (line == 3)
        printf("***");
    else if (line == 4)
        printf("*  ");
    else if (line == 5)
        printf("***");
}

void cout3(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("  *");
    else if (line == 3)
        printf("***");
    else if (line == 4)
        printf("  *");
    else if (line == 5)
        printf("***");
}

void cout4(int line) {
    if (line == 1)
        printf("* *");
    else if (line == 2)
        printf("* *");
    else if (line == 3)
        printf("***");
    else if (line == 4)
        printf("  *");
    else if (line == 5)
        printf("  *");
}

void cout5(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("*  ");
    else if (line == 3)
        printf("***");
    else if (line == 4)
        printf("  *");
    else if (line == 5)
        printf("***");
}

void cout6(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("*  ");
    else if (line == 3)
        printf("***");
    else if (line == 4)
        printf("* *");
    else if (line == 5)
        printf("***");
}

void cout7(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("  *");
    else if (line == 3)
        printf("  *");
    else if (line == 4)
        printf("  *");
    else if (line == 5)
        printf("  *");
}

void cout8(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("* *");
    else if (line == 3)
        printf("***");
    else if (line == 4)
        printf("* *");
    else if (line == 5)
        printf("***");
}

void cout9(int line) {
    if (line == 1)
        printf("***");
    else if (line == 2)
        printf("* *");
    else if (line == 3)
        printf("***");
    else if (line == 4)
        printf("  *");
    else if (line == 5)
        printf("***");
}

int main() {
    int A, B;
    scanf("%d%d", &A, &B);
    int res = A + B; // 数据保证 10 <= A + B < 100
    int num1 = res / 10;
    int num2 = res % 10;
    for (int j = 1; j <= 6; j++) {
        if (num1 == 0)
            cout0(j);
        else if (num1 == 1)
            cout1(j);
        else if (num1 == 2)
            cout2(j);
        else if (num1 == 3)
            cout3(j);
        else if (num1 == 4)
            cout4(j);
        else if (num1 == 5)
            cout5(j);
        else if (num1 == 6)
            cout6(j);
        else if (num1 == 7)
            cout7(j);
        else if (num1 == 8)
            cout8(j);
        else if (num1 == 9)
            cout9(j);
        putchar(' ');
        if (num2 == 0)
            cout0(j);
        else if (num2 == 1)
            cout1(j);
        else if (num2 == 2)
            cout2(j);
        else if (num2 == 3)
            cout3(j);
        else if (num2 == 4)
            cout4(j);
        else if (num2 == 5)
            cout5(j);
        else if (num2 == 6)
            cout6(j);
        else if (num2 == 7)
            cout7(j);
        else if (num2 == 8)
            cout8(j);
        else if (num2 == 9)
            cout9(j);
        putchar('\n');
    }
    return 0;
}

    大佬代码一（膜拜）

    #include<iostream>
using namespace std;
string x[15][10];
int ans[105],cou;
int main(){
	ios::sync_with_stdio(false);
	x[0][0]="***";
	x[0][1]="* *";
	x[0][2]="* *";
	x[0][3]="* *";
	x[0][4]="***";
	
	x[1][0]=" * ";
	x[1][1]=" * ";
	x[1][2]=" * ";
	x[1][3]=" * ";
	x[1][4]=" * ";
	
	x[2][0]="***";
	x[2][1]="  *";
	x[2][2]="***";
	x[2][3]="*  ";
	x[2][4]="***";
	
	x[3][0]="***";
	x[3][1]="  *";
	x[3][2]="***";
	x[3][3]="  *";
	x[3][4]="***";
	
	x[4][0]="* *";
	x[4][1]="* *";
	x[4][2]="***";
	x[4][3]="  *";
	x[4][4]="  *";
	
	x[5][0]="***";
	x[5][1]="*  ";
	x[5][2]="***";
	x[5][3]="  *";
	x[5][4]="***";
	
	x[6][0]="***";
	x[6][1]="*  ";
	x[6][2]="***";
	x[6][3]="* *";
	x[6][4]="***";
	
	x[7][0]="***";
	x[7][1]="  *";
	x[7][2]="  *";
	x[7][3]="  *";
	x[7][4]="  *";
	
	x[8][0]="***";
	x[8][1]="* *";
	x[8][2]="***";
	x[8][3]="* *";
	x[8][4]="***";
	
	x[9][0]="***";
	x[9][1]="* *";
	x[9][2]="***";
	x[9][3]="  *";
	x[9][4]="***";
	
	int a,b;
	cin>>a>>b;
	int xx=a+b,y=0,yy;
	while(xx){
		ans[cou++]=xx%10;
		xx/=10;
	}
	for(int i=0;i<5;i++){
		yy=y;
		for(int j=cou-1;j>=0;j--){
			cout<<x[ans[j]][i];
			if(j)cout<<' ';
		}
		cout<<endl;
	}
	
	
	
	return 0;
}

#include<iostream>
using namespace std;
char n0[6][6]={"***","* *","* *","* *","***"},n1[6][6]={" * "," * "," * "," * "," * "};
char n2[6][6]={"***","  *","***","*  ","***"},n3[6][6]={"***","  *","***","  *","***"};
char n4[6][6]={"* *","* *","***","  *","  *"},n5[6][6]={"***","*  ","***","  *","***"};
char n6[6][6]={"***","*  ","***","* *","***"},n7[6][6]={"***","  *","  *","  *","  *"};
char n8[6][6]={"***","* *","***","* *","***"},n9[6][6]={"***","* *","***","  *","***"};
int main()
{
	int a,b;
	cin>>a>>b;
	int c=a+b;
	char s[3];

	s[1]=c/10;
	s[2]=c%10;
	int flag=0;
			for(int p=0;p<5;p++)
			{
				for(int j=1;j<3;j++)
				{
					if(s[j]==1)cout<<n1[p];
					if(s[j]==2)cout<<n2[p];
					if(s[j]==3)cout<<n3[p];
					if(s[j]==4)cout<<n4[p];
					if(s[j]==5)cout<<n5[p];
					if(s[j]==6)cout<<n6[p];
					if(s[j]==7)cout<<n7[p];
					if(s[j]==8)cout<<n8[p];
					if(s[j]==9)cout<<n9[p];
					if(s[j]==0)cout<<n0[p];
                    if(j==1)cout<<" ";
				}
				cout<<"\n";
	}
	return 0;
}

    大佬代码二（继续膜拜）

    #include<iostream>
using namespace std;
char n0[6][6]={"***","* *","* *","* *","***"},n1[6][6]={" * "," * "," * "," * "," * "};
char n2[6][6]={"***","  *","***","*  ","***"},n3[6][6]={"***","  *","***","  *","***"};
char n4[6][6]={"* *","* *","***","  *","  *"},n5[6][6]={"***","*  ","***","  *","***"};
char n6[6][6]={"***","*  ","***","* *","***"},n7[6][6]={"***","  *","  *","  *","  *"};
char n8[6][6]={"***","* *","***","* *","***"},n9[6][6]={"***","* *","***","  *","***"};
int main()
{
	int a,b;
	cin>>a>>b;
	int c=a+b;
	char s[3];

	s[1]=c/10;
	s[2]=c%10;
	int flag=0;
			for(int p=0;p<5;p++)
			{
				for(int j=1;j<3;j++)
				{
					if(s[j]==1)cout<<n1[p];
					if(s[j]==2)cout<<n2[p];
					if(s[j]==3)cout<<n3[p];
					if(s[j]==4)cout<<n4[p];
					if(s[j]==5)cout<<n5[p];
					if(s[j]==6)cout<<n6[p];
					if(s[j]==7)cout<<n7[p];
					if(s[j]==8)cout<<n8[p];
					if(s[j]==9)cout<<n9[p];
					if(s[j]==0)cout<<n0[p];
                    if(j==1)cout<<" ";
				}
				cout<<"\n";
	}
	return 0;
}

    END 预祝学习愉快 /狗头

    • 0
      @ 2024-2-20 13:21:28

      打表并合理运用下标

      #include <bits/stdc++.h>
using namespace std;

#define DBG(x) cout << #x << "=" << x << endl

char g[10][5][5] = {
  {  // 0
    "***",
    "* *",
    "* *",
    "* *",
    "***",
  },
  {  // 1
    " * ",
    " * ",
    " * ",
    " * ",
    " * ",
  },
  {  // 2
    "***",
    "  *",
    "***",
    "*  ",
    "***",
  },
  {  // 3
    "***",
    "  *",
    "***",
    "  *",
    "***",
  },
  {  // 4
    "* *",
    "* *",
    "***",
    "  *",
    "  *",
  },
  {  // 5
    "***",
    "*  ",
    "***",
    "  *",
    "***",
  },
  {  // 6
    "***",
    "*  ",
    "***",
    "* *",
    "***",
  },
  {  // 7
    "***",
    "  *",
    "  *",
    "  *",
    "  *",
  },
  {  // 8
    "***",
    "* *",
    "***",
    "* *",
    "***",
  },
  {  // 9
    "***",
    "* *",
    "***",
    "  *",
    "***",
  }
};

int nums[3] = {0, 0, 0};

void solve() {
  int a, b;

  cin >> a >> b;
  int n = a + b;
  int t = 2;
  while (n) {
    nums[t--] = n % 10;
    n /= 10;
  }

  for (int r = 0; r < 5; r++) {
    for (int i = 0; i <= 2; i++) {
      if (i == 0 && nums[0] == 0) continue;
      if (i == 1 && nums[0] == 0 && nums[1] == 0) continue;
      int num = nums[i];
      cout << g[num][r] << (i == 2 ? "" : " ");
    }
    cout << endl;
  }
}

int main() {
  solve();

  return 0;
}
      • 0
        @ 2022-4-10 17:24:17 
        

        import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int end = sc.nextInt() + sc.nextInt();
        int n1, n2;
        n1 = end / 10;
        n2 = end % 10;
        String[][] stars = stars(n1, n2);
        printString(stars);
    }

    static String[][] copyT(String[][] a, String[][] b) {
        String[][] c = new String[5][7];
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j < 3; j++) {
                c[i][j] = a[i][j];
            }
        }
        for (int i = 0; i < 5; i++) {
            c[i][3] = " ";
        }
        for (int i = 0; i < 5; i++) {
            for (int j = 4; j < 7; j++) {
                c[i][j] = b[i][j - 4];
            }
        }
        return c;
    }

    static String[][] stars(int n1, int n2) {
        String[][] a = drawStars(n1);
        String[][] b = drawStars(n2);
        return copyT(a, b);
    }

    static String[][] drawStars(int n) {
        String[][] a = new String[5][3];
        reset(a);
        if (n == 0) {
            s0(a);
        } else if (n == 1) {
            s1(a);
        } else if (n == 2) {
            s2(a);
        } else if (n == 3) {
            s3(a);
        } else if (n == 4) {
            s4(a);
        } else if (n == 5) {
            s5(a);
        } else if (n == 6) {
            s6(a);
        } else if (n == 7) {
            s7(a);
        } else if (n == 8) {
            s8(a);
        } else if (n == 9) {
            s9(a);
        }
        return a;
    }

    static void reset(String[][] a) {
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j < 3; j++) {
                a[i][j] = "*";
            }
        }
    }

    static void s9(String[][] a) {
        s8(a);
        a[3][0] = " ";
    }

    static void s8(String[][] a) {
        s0(a);
        a[2][1] = "*";
    }

    static void s7(String[][] a) {
        for (int i = 1; i < 5; i++) {
            a[i][0] = " ";
            a[i][1] = " ";
        }
    }

    static void s6(String[][] a) {
        s5(a);
        a[3][0] = "*";
    }

    static void s5(String[][] a) {
        s3(a);
        a[1][0] = "*";
        a[1][2] = " ";
    }

    static void s4(String[][] a) {
        s9(a);
        a[0][1] = " ";
        a[4][1] = " ";
        a[4][0] = " ";
    }

    static void s3(String[][] a) {
        s2(a);
        a[3][2] = "*";
        a[3][0] = " ";
    }

    static void s2(String[][] a) {
        s0(a);
        a[2][1] = "*";
        a[1][0] = " ";
        a[3][2] = " ";
    }

    static void s1(String[][] a) {
        for (int i = 0; i < 5; i++) {
            a[i][0] = " ";
            a[i][2] = " ";
        }
    }

    static void s0(String[][] a) {
        a[1][1] = " ";
        a[2][1] = " ";
        a[3][1] = " ";
    }

    static void printString(String[][] a) {
        for (int i = 0; i < 5; i++) {
            for (int j = 0; j < 7; j++) {
                System.out.print(a[i][j]);
            }
            if (i != 4) {
                System.out.println();
            }
        }
    }
}
        • 1

        A + B 与繁星

        Information

        ID
        29
        Time
        1000ms
        Memory
        256MiB
        Difficulty
        3
        Tags
        # Submissions
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        Accepted
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