作业帮一个射击员打靶,请问连开10枪打90环的可能性有多少种?C语言递归实现,正确答案为92378,我的错在哪我的输出为184756,错在哪,#includeint num = 0;int score[10];voidfind (int remain,int n){if (remain < 0 || remain > 1
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/13 05:59:24
![一个射击员打靶,请问连开10枪打90环的可能性有多少种?C语言递归实现,正确答案为92378,我的错在哪我的输出为184756,错在哪,#includeint num = 0;int score[10];voidfind (int remain,int n){if (remain < 0 || remain > 1](/uploads/image/z/12627483-51-3.jpg?t=%E4%B8%80%E4%B8%AA%E5%B0%84%E5%87%BB%E5%91%98%E6%89%93%E9%9D%B6%2C%E8%AF%B7%E9%97%AE%E8%BF%9E%E5%BC%8010%E6%9E%AA%E6%89%9390%E7%8E%AF%E7%9A%84%E5%8F%AF%E8%83%BD%E6%80%A7%E6%9C%89%E5%A4%9A%E5%B0%91%E7%A7%8D%3FC%E8%AF%AD%E8%A8%80%E9%80%92%E5%BD%92%E5%AE%9E%E7%8E%B0%2C%E6%AD%A3%E7%A1%AE%E7%AD%94%E6%A1%88%E4%B8%BA92378%2C%E6%88%91%E7%9A%84%E9%94%99%E5%9C%A8%E5%93%AA%E6%88%91%E7%9A%84%E8%BE%93%E5%87%BA%E4%B8%BA184756%2C%E9%94%99%E5%9C%A8%E5%93%AA%2C%23includeint+num+%3D+0%3Bint+score%5B10%5D%3Bvoidfind+%28int+remain%2Cint+n%29%7Bif+%28remain+%3C+0+%7C%7C+remain+%3E+1)
一个射击员打靶,请问连开10枪打90环的可能性有多少种?C语言递归实现,正确答案为92378,我的错在哪我的输出为184756,错在哪,#includeint num = 0;int score[10];voidfind (int remain,int n){if (remain < 0 || remain > 1
一个射击员打靶,请问连开10枪打90环的可能性有多少种?C语言递归实现,正确答案为92378,我的错在哪
我的输出为184756,错在哪,
#include
int num = 0;
int score[10];
void
find (int remain,int n)
{
if (remain < 0 || remain > 10 * (n + 1))
return;
if (0 == n)
{
for (score[n] = 10; score[n] >= 0; score[n]--)
\x05{
\x05 if (score[n] < remain)
\x05 return;
\x05 else
\x05 num++;
\x05}
return;
}
for (score[n] = 10; score[n] >= 0; score[n]--)
{
find (remain - score[n],n - 1);
}
}
void
main ()
{
find (90,9);
printf ("num=%d\n",num);
}
一个射击员打靶,请问连开10枪打90环的可能性有多少种?C语言递归实现,正确答案为92378,我的错在哪我的输出为184756,错在哪,#includeint num = 0;int score[10];voidfind (int remain,int n){if (remain < 0 || remain > 1
if (0 == n)后面的for循环是用来判断最后一枪的环数的,只可能产生一种可能性,但是按你的写法,num++最多可以在一个循环中运行11次（remain=0时）
函数修改如下即可输出正确答案
#include
int num = 0;
int score[10];
void
find (int remain, int n)
{
if (remain < 0 || remain > 10 * (n + 1))
return;
if (0 == n)
{
for (score[n] = 10; score[n] >= 0; score[n]--)
{
if (score[n] < remain)
return;
else
num++;return;
}
return;
}
for (score[n] = 10; score[n] >= 0; score[n]--)
{
find (remain - score[n], n - 1);
}
}
void
main ()
{
find (90, 9);
printf ("num=%d\n", num);
}