1樓：聽不清啊
private sub command1_click()dim a(16) as integer, x as integerx = cint(text1.text)
if x >= 0 then a(16) = 0 else a(16) = 1
x = abs(x)
for i = 1 to 15
a(i) = x mod 2
x = x \ 2
next i
text2.text = ""
for i = 16 to 1 step -1text2.text = text2.text & a(i)next i
if a(16) = 0 then
text3.text = text2.texttext4.text = text2.textexit sub
end if
for i = 1 to 15
a(i) = 1 - a(i)
next i
text3.text = a(16)
for i = 15 to 1 step -1text3.text = text3.text & a(i)next i
text4.text = a(16)
a(1) = a(1) + 1
i = 1
while a(i) = 2 and i < 15a(i) = 0
a(i + 1) = a(i + 1) + 1i = i + 1
wend
for i = 15 to 1 step -1text4.text = text4.text & a(i)next i
end sub
一個整數n,用vb程式設計求它的原碼，補碼和反碼。
2樓：聽不清啊
private sub command1_click()dim a(16) as integer, x as integerx = cint(text1.text)
if x >= 0 then a(16) = 0 else a(16) = 1
x = abs(x)
for i = 1 to 15
a(i) = x mod 2
x = x \ 2
next i
text2.text = ""
for i = 16 to 1 step -1text2.text = text2.text & a(i)next i
if a(16) = 0 then
text3.text = text2.texttext4.text = text2.textexit sub
end if
for i = 1 to 15
a(i) = 1 - a(i)
next i
text3.text = a(16)
for i = 15 to 1 step -1text3.text = text3.text & a(i)next i
text4.text = a(16)
a(1) = a(1) + 1
i = 1
while a(i) = 2 and i < 15a(i) = 0
a(i + 1) = a(i + 1) + 1i = i + 1
wend
for i = 15 to 1 step -1text4.text = text4.text & a(i)next i
end sub
c語言程式題，任意輸入一個整數，編一個程式，使其輸出輸出相應的原碼、反碼及補碼。
3樓：great啦啦啦
程式如下：
#include
void main()}
4樓：
||**資料：
#include "stdio.h"
#include "limits.h"
void myout(unsigned n)int main(int argc,char *argv)printf("%d:\n",n);
printf("原碼專: ");
myout(n>=0 ? n : (~n+1)|~int_max);
printf("\n反碼: ");
myout(n>=0 ? n : ~(-n));
printf("\n補碼屬: ");
myout(n);
putchar('\n');
return 0;}
用c語言編一個程式,使給出一個數的原碼,求得反碼，補碼。
5樓：匿名使用者
#include
int main()
j = i-1;
ab[0] = af[0] = a[0];//符號位始終不變if(a[0] == 1)
}k = i;
for(; i > 0; i--)
af[i] = 1 - a[i];
for(i = k; i > 0; i--)//補碼從末位數起第一個不為0，以後均取反
ab[i] = 1 - a[i];
}else
}printf("對應的反碼是：\n");
for(i = 0; i <= j; i++)printf("%d", af[i]);
printf("\n");
printf("對應的補碼是：\n");
for(i = 0; i <= j; i++)printf("%d", ab[i]);
}在vc下編譯測試通過。另，本**未考慮輸入非法問題，所輸入的數字智慧由1和0組成
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