我最亲密的朋友正在上一门 EE 课程（我是他最后的希望：/），我大约 7 年前就了解 Java，但他的（大纲）最新的 EE 编程任务是使用 MIPS 程序集执行以下操作：
编写一个程序，接受两个正整数（m 和 n）并计算：
x= (m^n) - (1+2+3+…+n) * min(m,n)!
两个整数都应该大于零。我不允许使用任何 R 型算术指令（add、mult、sub）。相反，我要使用其他指令为其功能编写代码？？？？“您的程序应该在每次计算后继续获取 m 和 n 的新值，直到用户输入零，这将是您的程序的结束。”
我无权访问他以前的任何作业，并且尝试在不使用 (add, mult, sub) 的情况下一头扎进汇编语言对我来说效果不佳。
ece.ucdavis.edu/~vojin/CLASSES/EEC70/W2001/pr4.pdf 教授似乎正在使用他在加州大学戴维斯分校任教时的 ols 作业。
//edit 这是问题的 c++ 版本，它没有涵盖所有赋值的基础，但它是一个起点：
#include <iostream.h>
//x = (m^n) - (1+2+3+...+n) * ((min(m,n))!)
int m; //User Input
int n; //User Input
double answer; //Answer yo.
int findMin(int, int); //Takes 2 int inputs and outputs the smallest int.
int minFound; //Function output
double factorial(int); //Do eet.
double factOutput; //Function output
double sumN(int); //1+2+3+...+n
double sumFound; //Function output
double expMtoN(int, int); //m^n, float for number size,
double expFound; //Function output, float for number size,
int main(void)
{
cout << "Please enter a positive integer (m): ";
cin >> m;
//Escape if zero.
if ( m == 0)
{
cout << "User input for \"m\" is equal to zero; escape on zero." << endl;
return 0;
}
cout << "Please enter a positive integer (n): ";
cin >> n;
//Escape if zero.
if ( n == 0)
{
cout << "User input for \"n\" is equal to zero; escape on zero." << endl;
return 0;
}
expFound = expMtoN(m, n); //m^n
sumFound = sumN(n); //1+2+3+...+n
minFound = findMin(m, n); //Takes 2 int inputs and outputs the smallest int.
factOutput = factorial(minFound); //Factorial math for minFound (z!)
answer = expFound - sumFound * factOutput; //x = (m^n) - (1+2+3+...+n) * ((min(m,n))!)
cout << endl;
cout << m << " raised to the power of " << n << " is: " << expFound << endl;
cout << "Sum of " << n << " is: " << sumFound << endl;
cout << "Lowest number out of " << m << " and " << n << " is: " << minFound << endl;
cout << minFound << " factorial is: " << factOutput << endl;
cout << endl << "x = (m^n) - (1+2+3+...+n) * ((min(m,n))!)" << endl;
cout << "x = " << answer << endl;
}
//all temp variables below are confined to their respective functions.
//return functions output temp into variable from main.
double expMtoN(int userBase, int userExp)
{
double temp = 1; //Must establish 1 so you are not multiplying by zero.
for ( int i = 1; i <= userExp; i++ )
temp *= userBase;
return temp;
}
double sumN(int userN)
{
double temp = 0;
for ( int i = 1; i <= userN; i++ )
temp = temp + i;
return temp;
}
int findMin(int userM, int userN)
{
if( userM <= userN )
return userM;
else
return userN;
}
double factorial(int minFound)
{
double temp;
if ( minFound <= 1 )
return 1;
temp = minFound * factorial(minFound - 1);
return temp;
}
输入.s
;-----------------------------------------------------------------------------
;Subprogram call by symbol "InputUnsigned"
;expect the address of a zero-terminated prompt string in R1
;returns the read value in R1
;changes the contents of registers R1,R13,R14
;-----------------------------------------------------------------------------
.data
;*** Data for Read-Trap
ReadBuffer: .space 80
ReadPar: .word 0,ReadBuffer,80
;*** Data for Printf-Trap
PrintfPar: .space 4
SaveR2: .space 4
SaveR3: .space 4
SaveR4: .space 4
SaveR5: .space 4
.text
.global InputUnsigned
InputUnsigned:
;*** save register contents
sw SaveR2,r2
sw SaveR3,r3
sw SaveR4,r4
sw SaveR5,r5
;*** Prompt
sw PrintfPar,r1
addi r14,r0,PrintfPar
trap 5
;*** call Trap-3 to read line
addi r14,r0,ReadPar
trap 3
;*** determine value
addi r2,r0,ReadBuffer
addi r1,r0,0
addi r4,r0,10 ;Decimal system
Loop: ;*** reads digits to end of line
lbu r3,0(r2)
seqi r5,r3,10 ;LF -> Exit
bnez r5,Finish
subi r3,r3,48 ;´0´
multu r1,r1,r4 ;Shift decimal
add r1,r1,r3
addi r2,r2,1 ;increment pointer
j Loop
Finish: ;*** restore old register contents
lw r2,SaveR2
lw r3,SaveR3
lw r4,SaveR4
lw r5,SaveR5
jr r31 ; Return