Question

急求一个可被调用的C语言猜拳程序，不需要复杂功能，只要能进去，游戏，退出，就好

Answer 1

#include<stdio.h>
#include<math.h>
main()
{char c1;
int i,j;
do{
printf("Please input :");
scanf("%d",&i);
if(i<1||i>3){printf("Input error!\nPlease input again!\n");continue;}
j=rand()%3+1;
printf("%d\n",j);
if(i==j)printf("Ping Ju!\n");else
if((i-j)==1)printf("YOU WIN!\n");else
if(i==1&&j==3)printf("YOU WIN!\n");else
printf("YOU LOSE!\n");
printf("Replay?(Y/N)");
scanf("%c",&c1);
if(c1=='n'||c1=='N')break;
}while(c1=='y'||c1=='Y');
printf("Welcome to play again!\nAny key Quit...");
getch();putchar('\n');
}

追问

有问题啊

追答

报的错时什么？

追问

你的1.2.3哪个代表剪刀，哪个代表布，哪个代表石头

追答

#include
#include
#include
#include
main()
{
char c1;
int i,j;
do
{
printf("Please input :");
scanf("%d",&i);
if(i3)
{
printf("Input error!\nPlease input again!\n");
continue;
}
j=rand()%3+1;
printf("%d\n",j);
if(i==j)
printf("Ping Ju!\n");
else if((i-j)==1)
printf("YOU WIN!\n");
else if(i==1&&j==3)
printf("YOU WIN!\n");
else printf("YOU LOSE!\n");
printf("Replay?(Y/N)");
scanf("%c",&c1);
if(c1=='n'||c1=='N')
break;
}
while(c1=='y'||c1=='Y');
printf("Welcome to play again!\nAny key Quit...");
getch();putchar('\n');
}
石头 剪刀 布 1 2 3

Answer 2

怎么猜拳。

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石头剪刀布

追答

那还不简单，用c ++行不行

等下发给你

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？？

追答

我写好了没发

c++的