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c程序在循环程序时重复游戏的尝试

转载 作者:太空宇宙 更新时间:2023-11-04 07:53:11 26 4
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#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(void)
{
int num;
int guess = 0;
int response = 1;
int tries = 0;
srand(time(0));

while (response == 1)
{
printf("Please type your first guess.\n");
num = rand() % 1000 + 1;
while (guess != num)
{
scanf("%d", &guess);
tries = tries + 1;
if (guess == num)
{
printf("good job\n");
}
else if (guess > num)
{
printf("too high, try again\n");
}
else
{
printf("too low, try again\n");
}
}
printf("Guess Taken = %d\n", tries);
printf ("Would you like to play again? \n");
printf("Please type ( 1=yes, 2=no ) ");
scanf ("%d", &response);


}
system("pause");
return 0;
}

第二次运行程序“Guess Taken”的输出等于第一次和第二次的总和。如何解决问题?

例如,

第 1 次:9 次尝试

第二:8 次尝试

程序的输出显示了 17 次尝试而不是 8 次尝试。

最佳答案

 printf("I have a number between 1 and 1000.\n" 
"Can you guess my number?\n"
"Please type your first guess.\n");
num = rand() % 1000 + 1;
while(guess != num)

在进入 while 循环之前将 tries 设置为 0:

num = rand() % 1000 + 1;
tries = 0;
while(guess != num)

关于c程序在循环程序时重复游戏的尝试,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52503517/

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