检测四个开关S1～S4的状态，开关S1合上时，数码管显示1，开关S2合上时，数码管显示2，依次轮下去，当开关没有合上或合上的开关多于1个时，数码管均显示0。

这个问题可以通过编写一个简单的程序来解决。以下是一个可能的解决方案：

int s1 = 0;
int s2 = 0;
int s3 = 0;
int s4 = 0;

void setup() {
  pinMode(2, INPUT_PULLUP); // S1
  pinMode(3, INPUT_PULLUP); // S2
  pinMode(4, INPUT_PULLUP); // S3
  pinMode(5, INPUT_PULLUP); // S4
  pinMode(6, OUTPUT); // 数码管的 a 引脚
  pinMode(7, OUTPUT); // 数码管的 b 引脚
  pinMode(8, OUTPUT); // 数码管的 c 引脚
  pinMode(9, OUTPUT); // 数码管的 d 引脚
}

void loop() {
  s1 = digitalRead(2);
  s2 = digitalRead(3);
  s3 = digitalRead(4);
  s4 = digitalRead(5);

  if (s1 == LOW && s2 == HIGH && s3 == HIGH && s4 == HIGH) {
    displayNumber(1);
  } else if (s1 == HIGH && s2 == LOW && s3 == HIGH && s4 == HIGH) {
    displayNumber(2);
  } else if (s1 == HIGH && s2 == HIGH && s3 == LOW && s4 == HIGH) {
    displayNumber(3);
  } else if (s1 == HIGH && s2 == HIGH && s3 == HIGH && s4 == LOW) {
    displayNumber(4);
  } else {
    displayNumber(0);
  }
}

void displayNumber(int number) {
  switch (number) {
    case 0:
      digitalWrite(6, LOW);
      digitalWrite(7, LOW);
      digitalWrite(8, LOW);
      digitalWrite(9, LOW);
      break;
    case 1:
      digitalWrite(6, HIGH);
      digitalWrite(7, LOW);
      digitalWrite(8, LOW);
      digitalWrite(9, LOW);
      break;
    case 2:
      digitalWrite(6, LOW);
      digitalWrite(7, HIGH);
      digitalWrite(8, LOW);
      digitalWrite(9, LOW);
      break;
    case 3:
      digitalWrite(6, HIGH);
      digitalWrite(7, HIGH);
      digitalWrite(8, LOW);
      digitalWrite(9, LOW);
      break;
    case 4:
      digitalWrite(6, LOW);
      digitalWrite(7, LOW);
      digitalWrite(8, HIGH);
      digitalWrite(9, LOW);
      break;
    case 5:
      digitalWrite(6, HIGH);
      digitalWrite(7, LOW);
      digitalWrite(8, HIGH);
      digitalWrite(9, LOW);
      break;
    case 6:
      digitalWrite(6, LOW);
      digitalWrite(7, HIGH);
      digitalWrite(8, HIGH);
      digitalWrite(9, LOW);
      break;
    case 7:
      digitalWrite(6, HIGH);
      digitalWrite(7, HIGH);
      digitalWrite(8, HIGH);
      digitalWrite(9, LOW);
      break;
    case 8:
      digitalWrite(6, LOW);
      digitalWrite(7, LOW);
      digitalWrite(8, LOW);
      digitalWrite(9, HIGH);
      break;
    case 9:
      digitalWrite(6, HIGH);
      digitalWrite(7, LOW);
      digitalWrite(8, LOW);
      digitalWrite(9, HIGH);
      break;
  }
}

这个程序使用了 Arduino 板子上的数字输入和数字输出引脚，以及一个简单的 switch 语句来控制数码管的显示。当 S1～S4 中有一个合上时，程序会根据合上的开关来显示相应的数字，如果没有开关合上或者有多个开关合上，数码管会显示 0。