Lab 1/21 Switch練習
參考程式：
========================
#include
#include
#include
void main()
{
int i,count1=0,count2=0,count3=0,count4=0,count5=0,count6=0;
srand((unsigned)time(NULL));
for(i=1000;i>0;i--)
{
switch(rand()%6+1)
{
case 1:
count1++;
break;
case 2:
count2++;
break;
case 3:
count3++;
break;
case 4:
count4++;
break;
case 5:
count5++;
break;
case 6:
count6++;
break;
default:
break;
}
}
printf("擲骰子1000次出現點數3的次數為%d\n",count3);
}
========================
課堂練習：
請在原程式後面加上一個功能
使原本
printf("擲骰子1000次出現點數3的次數為%d\n",count5);
變更成可以選擇"所出現的點數"並只print那個點數出現的次數
posted by Aclose @ 9:41:00 AM
3 comments
3 Comments:
#include < stdio.h >
#include < stdlib.h >
#include < time.h >
int main()
{
int a,i,count1=0,count2=0,count3=0,count4=0,count5=0,count6=0;
srand((unsigned)time(NULL));
for(i=1000;i>0;i--)
{
switch(rand()%6+1)
{
case 1:
count1++;
break;
case 2:
count2++;
break;
case 3:
count3++;
break;
case 4:
count4++;
break;
case 5:
count5++;
break;
case 6:
count6++;
break;
default:
break;
}
}
printf("請輸入你想要知道次數的點數值(1~6)\n");
scanf("%d",&a);
switch(a)
{
case 1:
printf("擲骰子1000次出現點數1的次數為%d\n",count1);
break;
case 2:
printf("擲骰子1000次出現點數1的次數為%d\n",count2);
break;
case 3:
printf("擲骰子1000次出現點數1的次數為%d\n",count3);
break;
case 4:
printf("擲骰子1000次出現點數1的次數為%d\n",count4);
break;
case 5:
printf("擲骰子1000次出現點數1的次數為%d\n",count5);
break;
case 6:
printf("擲骰子1000次出現點數1的次數為%d\n",count6);
break;
default:
printf("您輸入的不是1~6喔!\n");
break;
}
return 0;
}
寫好了~
#include< stdio.h>
#include< stdlib.h>
#include< time.h>
void main()
{
int x,y,i,count1=0,count2=0,count3=0,count4=0,count5=0,count6=0;
srand((unsigned)time(NULL));
for(i=1000;i>0;i--)
{
switch(rand()%6+1)
{
case 1:
count1++;
break;
case 2:
count2++;
break;
case 3:
count3++;
break;
case 4:
count4++;
break;
case 5:
count5++;
break;
case 6:
count6++;
break;
default:
break;
}
}
printf("請輸入點數6~1\n");
scanf("%d",&x);
switch(x)
{
case 1:
printf("擲骰子1000次出現點數%d的次數為%d\n",x,count1++);
break;
case 2:
printf("擲骰子1000次出現點數%d的次數為%d\n",x,count1++);
break;
case 3:
printf("擲骰子1000次出現點數%d的次數為%d\n",x,count1++);
break;
case 4:
printf("擲骰子1000次出現點數%d的次數為%d\n",x,count4++);
break;
case 5:
printf("擲骰子1000次出現點數%d的次數為%d\n",x,count5++);
break;
case 6:
printf("擲骰子1000次出現點數%d的次數為%d\n",x,count6++);
break;
}
}
恩恩~~大致上就是這樣的概念
裕盛的小瑕疵是在print的部分沒有作變數的處理，應該只是忘記了
北極熊的就ok，不過你變數y沒有用到，可以不用宣告
其他如果有人後來才寫出來
一樣可以繼續回po
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