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;
}
}
}
}
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;
}
1.你的return 0呢。
2.你使用log()函数的时候有没有考虑过当number=0的时候，log（0）是非法的吗？