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59 lines
1007 B
Markdown
59 lines
1007 B
Markdown
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# 6-1 鸡兔同笼问题
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《孙子算经》记载：“今有雉兔同笼，上有三十五头，下有九十四足，问雉兔各几何？”
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#### 函数原型
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```c
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int ChickenRabbit(int *chicken, int *rabbit, int head, int foot);
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```
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说明：head 和 foot 为头和脚的数量，chicken 和 rabbit 为指示鸡和兔数量的指针。若问题有解，则将鸡和兔的数量保存到 chicken 和 rabbit 所指示的变量中，函数值为 1(真)；否则不改变 chicken 和 rabbit 所指示的变量，函数值为 0(假)。
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#### 裁判程序
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```c
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#include <stdio.h>
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int ChickenRabbit(int *chicken, int *rabbit, int head, int foot);
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int main()
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{
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int h, f, c, r;
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scanf("%d%d", &h, &f);
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if (ChickenRabbit(&c, &r, h, f))
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{
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printf("%d %d\n", c, r);
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}
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else
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{
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puts("None");
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}
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return 0;
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}
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/* 你的提交代码将被嵌在这里 */
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```
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#### 输入样例1
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```in
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35 94
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```
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#### 输出样例1
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```out
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23 12
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```
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#### 输入样例2
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```in
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30 71
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```
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#### 输出样例2
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```out
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None
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```
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