题解

这道题的解法需要我们结合我们解方程的方法：即首先把未知数移动到等式左侧，常数移动到等式右侧，最后除以下。 时间仓促，可能不太全面，凑合看看，欢迎Copy

#include <bits/stdc++.h>
bool flag, havenum;
char s[100010], ansx;
int son, mother, pre, lor = 1, equal, len;
int main()
{
    scanf("%s", s + 1); // 由于这里用的是字符数组
    len = strlen(s + 1);
    s[0] = '+';
    s[len + 1] = '+';
    len++;
    for (int i = 1; i <= len; i++)
    {
        if (s[i] >= 'a' && s[i] <= 'z') ansx = s[i]; // 记录未知数名
        if (s[i] == '=' || s[i] == '-' || s[i] == '+')
        {
            havenum = 0;
            if (i == 1 || i - 1 == equal)
            {
                pre = i;
                continue;
            }
            int x = 0;
            if (isdigit(s[i - 1])) flag = 0;
            else flag = 1;
            for (int j = pre + 1; j <= i - flag - 1; j++)
            {
                x = (x << 1) + (x << 3) + (s[j] ^ '0'); // 这里其实是 x * 10 + (s[j] - '0')
                havenum = 1;
            }
            if (havenum == 0) x = 1;
            if (flag == 0) son += x * lor * -1 * (s[pre] - 44) * -1; // 常数
            else mother += x * lor * (s[pre]- 44) * -1; // 未知数的系数
            pre = i;
            if (s[i] == '=') lor = -1, s[i] = '+', equal = i; // 过等号要变号,lor = -1
        }
    }
    double ansy = double(son) / double(mother); // 除出未知数
    if (fabs(ansy) - 0.0 < 0.000001) // 误差归0
        ansy = 0.0;
    printf("%c=%.3lf", ansx, ansy); // .3f输出
    return 0;
}

$AC$ $Code$

#include <bits/stdc++.h>
bool flag, havenum;
char s[100010], ansx;
int son, mother, pre, lor = 1, equal, len;
int main()
{
    scanf("%s", s + 1);
    len = strlen(s + 1);
    s[0] = '+';
    s[len + 1] = '+';
    len++;
    for (int i = 1; i <= len; i++)
    {
        if (s[i] >= 'a' && s[i] <= 'z') ansx = s[i];
        if (s[i] == '=' || s[i] == '-' || s[i] == '+')
        {
            havenum = 0;
            if (i == 1 || i - 1 == equal)
            {
                pre = i;
                continue;
            }
            int x = 0;
            if (isdigit(s[i - 1])) flag = 0;
            else flag = 1;
            for (int j = pre + 1; j <= i - flag - 1; j++)
            {
                x = (x << 1) + (x << 3) + (s[j] ^ '0');
                havenum = 1;
            }
            if (havenum == 0) x = 1;
            if (flag == 0) son += x * lor * -1 * (s[pre] - 44) * -1;
            else mother += x * lor * (s[pre]- 44) * -1;
            pre = i;
            if (s[i] == '=') lor = -1, s[i] = '+', equal = i;
        }
    }
    double ansy = double(son) / double(mother);
    if (fabs(ansy) - 0.0 < 0.000001)
        ansy = 0.0;
    printf("%c=%.3lf", ansx, ansy);
    return 0;
}

个位大佬帮忙看看，思路就是模拟，咋改改

#include <bits/stdc++.h>
using namespace std;
char s[100010];
char x;
int a;
double ans;
double sum;
double sum1;
int main(){
	cin>>s;
	int len=strlen(s);
	for (int i=0;i<len;i++){
	    if (s[i]>='a'&&s[i]<='z'){
	    	x=s[i];
	    	break;
		}
		if (s[i]=='='){
			a=i;
		}	
	}
	for (int i=0;i<len;i++){
		if (s[i]==x){
			ans=0;
			for (int j=i-1;j>=0;j--){
				if (s[j]>='0'&&s[j]<='9'){
					double u=s[j]-'0';
					ans=ans*10+u;
					if (j==0){
						if (s[j]>='0'&&s[j]<='9'){
							sum1+=ans;
						}
					}
				}
				else{
					if (s[j]=='+'){
						if (j<a){
							sum1+=ans;
						}
						else{
							sum1-=ans;
						}
					}
					if (s[j]=='-'){
						if (j<a){
							sum1-=ans;
						}
						else{
							sum1+=ans;
						}
					}
					break;
				}
			}
		}
		else if (s[i]>='0'&&s[i]<='9'){
			double n=0;
			for (int j=i-1;j>=0;j--){
				if (s[j]=='-'){
					if (j<a){
						sum+=n;
					    break;
					}
					else{
						sum-=n;
					    break;
					}
				}
				else if (s[j]=='+'){
					if (j<a){
						sum-=n;
					    break;
					}
					else{
						sum+=n;
					    break;
					}
				}
				else if (s[j]=='='){
					sum+=n;
					break;
				}
				else{
					double o=s[j]-'0';
					n=n*10+o;
				}
			}
		}
	}
	cout<<x<<"=";
	if (sum==0){
		cout<<"0.000";
	}
	else{
		printf("%.3lf\n",sum*1.0/sum1);
	}
	return 0;
}

核心思路：将未知数左移，常数右移，最后相除

#include <cstdio>

int val1 , val2 , n ;
//val1存储常数，val2存储未知数系数
bool left = true , tf = true;
//left是否在左 tf是正负数
//对于常数左减右加，未知数相反
char chr , chx;
//chr存储输入的字符，chx存储未知数字符

int main(){
	chr = getchar() ;
	while(
		('0' <= chr && chr <= '9') ||
		('a' <= chr && chr <= 'z') || 
		chr == '+' || chr == '-' || chr == '='
	){
        //string a ;
        //cin >> a ;
        //for(int i = 0 ; i < a.length ; i ++)
        //也可以
		if(chr == '=')
			val1 -= n * (tf ? 1 : -1) , n = 0 , left = false , tf = 1;
            //判定'=' 右边(val1)增加n，清空n，接下来就是右边
            //默认符号正号，tf = 1
            //这里一定左边，所以不用判断左右
		else if('0' <= chr && chr <= '9')
			n = (n << 3) + (n << 1) + chr - '0' ;
            //正常的数，这里与正负号无关
            //只需要+ 10 + 新个位数
		else if('a' <= chr && chr <= 'z')
			val2 += (n == 0 ? 1 : n) * 
			(left ? 1 : -1) * (tf ? 1 : -1), 
			n = 0 , chx = chr;
            //未知数
            //默认为1而不是0
            //左加右减，判断正负
            //清空n , 未知数变量修改
        //这时清空了n，若接下运算符或空
        //加减0等于寂寞
		else if(chr == '+')
			val1 += n * (left ? -1 : 1) * (tf ? 1 : -1) , 
			tf = 1 , n = 0 ;
            //常数
            //左减右加，判断正负
            //下一个是正数，清空n
		else if(chr == '-')
			val1 += n * (left ? -1 : 1) * (tf ? 1 : -1) , 
			tf = 0 , n = 0 ;
            //常数
            //左减右加，判断正负
            //下一个是负数，清空n
		chr = getchar() ;
	}
	val1 += n * (left ? -1 : 1) * (tf ? 1 : -1) ;
    //结尾剩余的加上
	if(val1 == 0)
		printf("%c=0.000" , chx) ;
        //特判0 防止出现-0.000
	else
		printf("%c=%.3lf" , chx ,  (1.0 * val1) / val2) ;
        //正常输出
	return 0 ;
}