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multiplication.cpp
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//
// Created by andy on 2022/7/18.
//
#include "multiplication.h"
#include "addition.h"
//#define DEBUG
#ifdef DEBUG
#define _getch() ;
#endif
Coordinate multiplicationRhymeDisplayLeftTop={885,150},multiplicationRhymeDisplayRightBottom{1350,183};
//显示表达式计算的结果
void drawExpressionOfMultiplication()
{
setFontSizeTo32();
double facNumber = atof(ochar_1operand);
double mulNumber = atof(ochar_2operand);
stringGenerator<<facNumber<<"*"<<mulNumber<<"="<<facNumber*mulNumber;
strcpy(strInfo,stringGenerator.str().c_str());
drawExpression(strInfo);
stringGenerator.str("");
setFontSizeTo16();
}
//绘制一个数的盘式
void drawNumOnAbacusOfMultiplication(Num *sa)
{
cleardevice(); //清空屏幕内容
drawExpressionOfMultiplication(); // 绘制列式计算的结果
drawAbacus(sa,AbacusParams()); //绘制算盘
// drawMnemonicRhymeOfMultiplication(); //绘制
}
void drawNumOnAbacusOfMultiplicationPureVersion(Num *sa)
{
AbacusParams param;
clearAbacus(param);
drawAbacus(sa,AbacusParams()); //绘制算盘
}
/**
* 模拟大九九口诀表返回乘积
* @param fac
* @param mul
* @param indexOfFirstDigit 积的第一个数字所在的挡位
* @param product
*/
void lookUpMultiplicationTable(int _product, int indexOfFirstDigit, Num* product){
if(_product<10){
setNumToAbacusIndexVersion(_product, product, indexOfFirstDigit + 1); //0不用放置，直接放置第二位
}else{
setNumToAbacusIndexVersion(_product / 10, product, indexOfFirstDigit); //放置第一位
setNumToAbacusIndexVersion(_product % 10, product, indexOfFirstDigit + 1); //放置第二位
}
}
void lookUpMultiplicationTableDivisionVersion(int _product, int indexOfFirstDigit, Num* product){
if(_product<10){
setNumToAbacusIndexVersion(_product, product, indexOfFirstDigit); //0不用放置，直接放置第二位
}else{
setNumToAbacusIndexVersion(_product / 10, product, indexOfFirstDigit); //放置第一位
setNumToAbacusIndexVersion(_product % 10, product, indexOfFirstDigit + 1); //放置第二位
}
}
/**
*
* @param result
* @param integerLen1 被乘数位数
* @param integerLen2 乘数位数（位数少的那一个）
*/
void simulateMultiplication(Num* result, int integerLen1, int integerLen2){
//定位
// int productDigitNum = oneToNumber(&num_1operand[PLACES_NUM - integerLen1]) *
// oneToNumber(&num_2operand[PLACES_NUM - integerLen2]) < 10 ? integerLen1 + integerLen2 - 1 : integerLen1 + integerLen2; //积的位数
int productDigitNum = integerLen1 + integerLen2; //积的位数
productDigitNum+=2; //加上两位小数的位置
Num* fac = num_1operand;
Num* mul = num_2operand;
Num* currentFac; //被乘数当前位
Num* currentMul; //乘数当前位
int outerAccumulationPointer=PLACES_NUM-productDigitNum; //1*n位乘积累加的错位指针
int innerAccumulationPointer; // 1位乘法累加的错位指针
Num product[PLACES_NUM]; //1位乘法之积
int _product;
stringstream ss; //yet another string builder
char oneToNExpressionStrInfo[20];
//乘法
for (int i = 0; i < integerLen2 + 2; ++i,outerAccumulationPointer++) { // 1*n位乘法的积错位相加
currentMul = &mul[PLACES_NUM - (integerLen2 + 2) + i];
if(oneToNumber(currentMul)==0){ //乘数当前位为零
if(i==integerLen2){ //当前位为十分位，没必要继续算法了，out
break;
}else{ //当前位在整数位，跳过当前循环加速计算
continue;
}
}
innerAccumulationPointer = outerAccumulationPointer;
ss << ochar_1operand << "*" << oneToNumber(currentMul);
strcpy(oneToNExpressionStrInfo,ss.str().c_str());
ss.clear();
ss.str("");
for (int j = 0; j < integerLen1 + 2; ++j,innerAccumulationPointer++) { // 1位乘法的积错位相加 5*7821
currentFac = &fac[PLACES_NUM - (integerLen1 + 2) + j];
//得1位乘法之积
_product = oneToNumber(currentFac)*oneToNumber(currentMul);
if(_product==0){ //被乘数当前位为0，不用算了
continue;
}
lookUpMultiplicationTable(_product, innerAccumulationPointer, product);
//显示累加之积
stringGenerator<<"错位累加"<<oneToNumber(currentFac)<<"*"<<oneToNumber(currentMul)<<"=";
if(_product<10){
stringGenerator<<"0"<<_product;
}
else{
stringGenerator<<_product;
}
strcpy(strInfo,stringGenerator.str().c_str());
drawMultiplicationRhymeProduct(oneToNExpressionStrInfo,strInfo);
stringGenerator.clear();
stringGenerator.str("");
//将积累加到结果中
_getch();
for (int k = PLACES_NUM-productDigitNum; k < PLACES_NUM; k++){
if(oneToNumber(&product[k]) != 0){
simulateAdditionPureVersion(result, product, k);
}
}
clearAbacus(product);
//show result
drawNumOnAbacusOfMultiplication(result);
}
}
}
//显示1*n位乘法运算提示
void draw1toNExpression(const char* str){
setFontSizeTo32();
RECT r1 = {multiplicationRhymeDisplayLeftTop.x, multiplicationRhymeDisplayLeftTop.y,
multiplicationRhymeDisplayRightBottom.x, multiplicationRhymeDisplayRightBottom.y};
drawtext(str, &r1, DT_WORDBREAK | DT_VCENTER | DT_LEFT);
setFontSizeTo16();
}
//显示1位乘法运算提示
void drawMultiplicationRhymeProduct(const char* oneToNExpressionStrInfo,const char *str) {
draw1toNExpression(oneToNExpressionStrInfo);
drawRules(str);
}
void simulateMultiplicationPureVersion(Num *fac, Num* mul,Num *result, int integerLen1, int integerLen2) {
int productDigitNum = integerLen1 + integerLen2; //积的位数
productDigitNum+=2; //加上两位小数的位置
Num* currentFac; //被乘数当前位
Num* currentMul; //乘数当前位
int outerAccumulationPointer=PLACES_NUM-productDigitNum; //1*n位乘积累加的错位指针
int innerAccumulationPointer; // 1位乘法累加的错位指针
Num product[PLACES_NUM]; //1位乘法之积
int _product;
stringstream ss; //yet another string builder
//乘法
for (int i = 0; i < integerLen2 + 2; ++i,outerAccumulationPointer++) { // 1*n位乘法的积错位相加
currentMul = &mul[PLACES_NUM - (integerLen2 + 2) + i];
if(oneToNumber(currentMul)==0){ //乘数当前位为零
if(i==integerLen2){ //当前位为十分位，没必要继续算法了，out
break;
}else{ //当前位在整数位，跳过当前循环加速计算
continue;
}
}
innerAccumulationPointer = outerAccumulationPointer;
for (int j = 0; j < integerLen1 + 2; ++j,innerAccumulationPointer++) { // 1位乘法的积错位相加 5*7821
currentFac = &fac[PLACES_NUM - (integerLen1 + 2) + j];
//得1位乘法之积
_product = oneToNumber(currentFac)*oneToNumber(currentMul);
if(_product==0){ //被乘数当前位为0，不用算了
continue;
}
lookUpMultiplicationTable(_product, innerAccumulationPointer, product);
//将积累加到结果中
_getch();
for (int k = PLACES_NUM-productDigitNum; k < PLACES_NUM; k++){
if(oneToNumber(&product[k]) != 0){
simulateAdditionPureVersion(result, product, k);
}
}
clearAbacus(product);
drawNumOnAbacusOfMultiplicationPureVersion(result);
}
}
}
//# pragma warning (disable:4819)