第四周 程序跳转

1.接力break

输入x元，输出用多少个1角，2角，5角来凑成x元

#include<stdio.h>

int main()
{
  int x;
  int one, two, five;

  scanf("%d", &x);
  for ( one = 1; one < x*10; one++) {
    for (two = 1; two < x*10/2; two++) {
      for ( five = 1; five < x*10/5; five++) {
        if (one + two*2 + five*5 == x*10) {
          printf("%d one + %d two + %d five = %d\n", one, two, five, x);
        }
      }
    }
  }

  return 0;
}

输入2元，输出：

1 one + 2 two + 3 five = 2
1 one + 7 two + 1 five = 2
2 one + 4 two + 2 five = 2
3 one + 1 two + 3 five = 2
3 one + 6 two + 1 five = 2
4 one + 3 two + 2 five = 2
5 one + 5 two + 1 five = 2
6 one + 2 two + 2 five = 2
7 one + 4 two + 1 five = 2
8 one + 1 two + 2 five = 2
9 one + 3 two + 1 five = 2
11 one + 2 two + 1 five = 2
13 one + 1 two + 1 five = 2

假如我只需要一种方法就可以了， 怎么输出一种方案后就停止呢？

如果在13行后，加条break；程序没有什么变化。为什么会这样呢？break只会跳出最内层的for语句。

那么在13，15，16行都加break，假如只这样的话，内层结束后，总会执行break。

所以要加个条件限制，设置一个变量 int exit = 0;　　然后在　break  前加 exit = 1;

#include<stdio.h>

int main()
{
  int x;
  int one, two, five;
  int exit = 0;

  scanf("%d", &x);
  for ( one = 1; one < x*10; one++) {
    for (two = 1; two < x*10/2; two++) {
      for ( five = 1; five < x*10/5; five++) {
        if (one + two*2 + five*5 == x*10) {
          printf("%d one + %d two + %d five = %d\n", one, two, five, x);
          exit = 1;
          break;
        }
      } 
      if (exit) break;
    }
    if (exit) break;
  }

  return 0;
}

这样的结果就是

1 one + 2 two + 3 five = 2

2.goto

或者用goto语句也可以达到这样的效果：

#include<stdio.h>

int main()
{
  int x;
  int one, two, five;

  scanf("%d", &x);
  for ( one = 1; one < x*10; one++) {
    for (two = 1; two < x*10/2; two++) {
      for ( five = 1; five < x*10/5; five++) {
        if (one + two*2 + five*5 == x*10) {
          printf("%d one + %d two + %d five = %d\n", one, two, five, x);
          goto out;
        }
      } 
    }
  }
  out:
  return 0;
}

goto特别合适这样的场合，跳出多重循环。别的场合使用的时候要慎重。