内存中有10个分布在0至100内的正整数， 求小于60的数的个数num1，大于或等于60且小于80的数的个数num2，大于或等于80且小于100的数的个数num3

C语言描述该程序流程：

#include <stdio.h>
int main()
{
	int a[]={1, 20, 95, 32, 60, 70, 87, 56, 93, 99};
	int num1=0,num2=0,num3=0;
	for(int i=0;i<10;i++)
	{
		if(a[i]<60)
		{
			num1++;
		}
		else if(a[i]<80)
		{
			num2++;
		}
		else
		{
			num3++;
		}
	}
	printf("%d\n%d\n%d",num1,num2,num3);
}

汇编语言：

include irvine32.inc
.data
numbers dword 1, 20, 95, 32, 60, 70, 87, 56, 93, 99
num1 dword 0
num2 dword 0
num3 dword 0
count dword 10 ;需比较十次
.code
main proc
mov esi,offset numbers

L1:
cmp count,0
jz output
mov ebx,[esi]
cmp ebx,60
jl smallerthan60
cmp ebx,80
jl smallerthan80
inc num3
jmp nextnum

smallerthan60:
inc num1
jmp nextnum

smallerthan80:
inc num2
jmp nextnum

nextnum:
dec count
add esi,4
jmp L1

output:
mov eax,num1
call writeint
call crlf
mov eax,num2
call writeint
call crlf
mov eax,num3
call writeint

main endp
end main

运行结果：