c++题目，可以指点一下这道题吗？想了特别久，样例已经实现了但是交的时候是runtime error

Question

Description

 

科学记数法是一种记录大数字的简单方法，但是无聊的栗酱修改了规则，这种记数方式可以把大数字用形如 [+/-][B].[C]E[+/-][D] 的字符串表示，其中B,C,DB,C,D的长度分别满足：|B|\le300,|C|\le300,|D|\le5∣B∣≤300,∣C∣≤300,∣D∣≤5，BB保证无前导零，且输入都满足以上格式。

但是这种表示形式过于复杂，栗酱已经不知道如何把它还原成原来的数了，所以聪明的你帮帮栗酱吧。

Input

 

第一行输入数据组数TT。

接下来有TT行，每行输入一行非空字符串ss。

T \le 20T≤20

|s|\le300∣s∣≤300

Output

 

每组数据一行，输出还原后的数。

Sample Input 1 

3
+123.4E+03
-123.40E-4
+123.4000E+2

Sample Output 1

123400
-0.012340
12340.00

#include<iostream>
#include<cmath>
#include<string>
using namespace std;
int main()
{
	int i, j, k,l;
	int T;
	cin >> T;
	for (i = 0; i < T; i++) {
		string s;
		string n1 = "";
		int n = 0;
		char a, b;
		cin >> s;
		for (j = 0; j < s.length(); j++) {
			if (s[j] == '+' || s[j] == '-') {
				a = s[j]; continue;
			}
			if (s[j] == '.') {
				l = j; continue;
			}
			if (s[j] == 'E')break;
			n1 += s[j];
		}
		for (k = j + 1; k < s.length(); k++) {
			if (s[k] == '+' || s[k] == '-')b = s[k];
			else n = n * 10 + s[k] - '0';
		}
		int ci;
		ci = 0-(j - l - 1);
		int sum;
		if (b =='-')sum = ci - n;
		else sum = ci + n;
		double number = stoi(n1);
		int d = log10(number) + 1;
		int t = abs(sum);
		cout << a;
		if (sum >= 0) {
			number = number * pow(10, sum);
			cout<< number << endl;
		}
		else {
			if (t >= d) {
				cout << "0.";
				for (int i = 0; i < t - d; i++)cout << '0';
				cout << number << endl;
			}
			else {
				cout << n1.substr(0,d-t) <<'.'<< n1.substr(d - t) <<endl;
			}
		}
	


		
	}
		

}

Answer 1

代码如下，如有帮助，请采纳一下，谢谢。

#include <stdio.h>
#include <string>

using namespace std;

string trans(char* buf)
{
	string s = buf;
	int len = strlen(buf);
	int indexe = s.find('E');
	char c = buf[indexe + 1]; //正还是负

	//E后面的数字
	string strnmb = s.substr(indexe+2,s.length() - indexe -1);
	int nb = atoi(strnmb.c_str()); 

	string strout = "";
	//数据部分
	string dd = "";
	if (buf[0] == '+' )
	{
		dd = s.substr(1,indexe -1);
	}
	else if (buf[0] == '-')
	{
		dd = s.substr(1,indexe -1);
		strout = "-";
	}
	else
		dd = s.substr(0,indexe);
	
	//查找数据部分小数点的位置
	int indexdot = dd.find('.');
	if (indexdot > 0 && indexdot < dd.length())
	{
		string zs = dd.substr(0,indexdot);
		string xs = dd.substr(indexdot+1,dd.length() - indexdot -1);
		//有小数点
		if(c == '+') //小数点右移动
		{
			string sm = "";
			if (xs.length() > nb)
			{
				sm = xs.substr(0,nb);
				strout = strout + zs + sm + "." + xs.substr(nb-1,xs.length() -nb);
			}else if (xs.length() == nb)
			{
				sm = xs;
				strout = strout + zs + sm;
			}else
			{
				sm = xs;
				while(sm.length()< nb)
					sm += "0";
				strout = strout + zs + sm;
			}
		}else if (c == '-') //小数点左移
		{
			string sm = "";
			if (zs.length() > nb)
			{
				sm = zs.substr(0,zs.length() - nb) + "." + zs.substr(zs.length() - nb-1,nb);
				strout = strout + sm + xs;
			}else if (zs.length() == nb)
			{
				sm = "0."+ zs;
				strout = strout + sm + xs;
			}else
			{
				sm = zs;
				while(sm.length() < nb)
					sm = "0" + sm;
				sm = "0." + sm;
				strout = strout + sm + xs;
			}
		}
	}else
	{
		//没有小数点
		if (c == '+') //右移
		{
			string sm = "";
			while (sm.length() < nb)
				sm += "0";
			strout = strout + sm;
		}else if (c == '-')
		{
			string sm = "";
			if (dd.length() > nb)
			{
				sm = dd.substr(0,dd.length() - nb) + "." + dd.substr(dd.length() - nb -1,nb);
				strout = strout + sm;
			}else if (dd.length() == nb)
			{
				strout = strout + "0." + dd;
			}else
			{
				while(dd.length() < nb)
					dd = "0"+dd;
				strout = strout + "0." + dd;
			}
		}

	}



	return strout;
}


int main()
{
	int nmb = 0;
	scanf("%d",&nmb);
	char buf[20][20] = {0};
	for (int i = 0; i < nmb;i++)
	{
		scanf("%s",buf[i]);
	}

	for (int i = 0; i < nmb; i++)
	{
		printf("%s\n",trans(buf[i]).c_str());
	}
	



	getchar();
	getchar();
	return 0;
}