5-1 科学计数法的值
科学计数法是一种数学专用术语。将一个数表示成 a×10的n次幂的形式，其中1≤|a|<10，n为整数，这种记数方法叫科学计数法。例如920000可以表示为9.2*10^5。现在需要对输入的字符串进行分离，自动识别该科学计数法中的a和幂次，计算其表征的具体数值并输出该值。例如，对于输入的复数字符串“9.210^5”，输出 The actual value for 9.210^5 is 920000
注意：
- 1、每组测试数据仅包括一个用于科学计数法的字符串。
- 2、输入字符串保证合法。
- 3、字符串长度不超过1000
- 4、幂次不超过200
输入示例：
9.2*10^5
输出示例：
The actual value for 9.2*10^5 is 920000
#include<iostream>
#include<sstream>
#include<string>
#include<cstdio>
using namespace std;
int main()
{
string str;
while(cin >> str)
{
string ans;
bool flag,IsSign,IsMinus;
int i;
int len = str.size();
string::size_type loc = str.find("^");
if (str[loc + 1] != '-')
{
IsSign = false;
}
else
{
IsSign = true;
}
if (!IsSign)
{
int intcnt = 0;
for (i = 0; str[i] != '*' && str[i] != '.'; i++)
{
ans += str[i];
intcnt++;
}
if (str[i] == '.')
{
i++;
}
int cnt = 0;
for (; str[i] != '*'; i++)
{
ans += str[i];
cnt++;
}
int record;
string tmp;
stringstream ss;
if (loc == string::npos)
{
record = 1;
}
else
{
for (i = loc+1; i < len; i++)
{
tmp += str[i];
}
ss << tmp;
ss >> record;
}
if (record < cnt)
{
flag = false;
}
else
{
flag = true;
}
int len1 = ans.size();
if (flag)
{
cout << "The actual value for " << str << " is " << ans;
for (i = record - cnt; i > 0; i--)
{
cout << "0";
}
cout << endl;
}
else
{
cout << "The actual value for " << str << " is ";
for (i = 0; i < intcnt + record; i++)
{
cout << ans[i];
}
cout << ".";
for (i = intcnt + record; i < len1; i++)
{
cout << ans[i];
}
cout << endl;
}
}
else
{
if (str[0] == '-')
{
str.erase(0,1);
IsMinus = true;
}
else
{
IsMinus = false;
}
len = str.size();
int intcnt = 0;
for (i = 0; str[i] != '*' && str[i] != '.'; i++)
{
ans += str[i];
intcnt++;
}
if (str[i] == '.')
{
i++;
}
for (; str[i] != '*'; i++)
{
ans += str[i];
}
int record;
string tmp;
stringstream ss;
if (loc == string::npos)
{
record = 1;
}
else
{
if (IsMinus)
{
for (i = loc + 1; i < len; i++)
{
tmp += str[i];
}
}
else
{
for (i = loc + 2;i < len;i++)
{
tmp += str[i];
}
}
ss << tmp;
ss >> record;
}
cout << "The actual value for " << str << " is ";
if (IsMinus)
{
cout << "-";
}
cout << "0.";
for (i = 0; i < record - intcnt; i++)
{
cout << "0";
}
cout << ans << endl;
}
}
return 0;
}
5-3 不能用循环是一件多么悲伤的事
下面是一个算到10的加法表：
0 + 0 = 0 0 + 1 = 1 0 + 2 = 2 0 + 3 = 3 0 + 4 = 4 0 + 5 = 5 0 + 6 = 6 0 + 7 = 7 0 + 8 = 8 0 + 9 = 9 0 +10 = 10
1 + 0 = 1 1 + 1 = 2 1 + 2 = 3 1 + 3 = 4 1 + 4 = 5 1 + 5 = 6 1 + 6 = 7 1 + 7 = 8 1 + 8 = 9 1 + 9 = 10
2 + 0 = 2 2 + 1 = 3 2 + 2 = 4 2 + 3 = 5 2 + 4 = 6 2 + 5 = 7 2 + 6 = 8 2 + 7 = 9 2 + 8 = 10
3 + 0 = 3 3 + 1 = 4 3 + 2 = 5 3 + 3 = 6 3 + 4 = 7 3 + 5 = 8 3 + 6 = 9 3 + 7 = 10
4 + 0 = 4 4 + 1 = 5 4 + 2 = 6 4 + 3 = 7 4 + 4 = 8 4 + 5 = 9 4 + 6 = 10
5 + 0 = 5 5 + 1 = 6 5 + 2 = 7 5 + 3 = 8 5 + 4 = 9 5 + 5 = 10
6 + 0 = 6 6 + 1 = 7 6 + 2 = 8 6 + 3 = 9 6 + 4 = 10
7 + 0 = 7 7 + 1 = 8 7 + 2 = 9 7 + 3 = 10
8 + 0 = 8 8 + 1 = 9 8 + 2 = 10
9 + 0 = 9 9 + 1 = 10
10+ 0 = 10
本题目要求读入1个整数，输出加法表，每一行都算到结果为输入的整数为止。不允许使用循环，不允许使用循环，不允许使用循环。重要的事情说三遍（括弧笑）
输入格式:
在一行中给出一个正整数N（0≤N≤99）。
输出格式:
按照示例的格式输出左上三角N+M的表，行列都从0开始。
加号左边数字占2位、左对齐；
加号右边数字占2位、右对齐；
结果数字占2位，左对齐。
等号两边各一个空格。
两个式子之间加一个空格（行末的空格不用去掉）
输入样例:
5
输出样例:
0 + 0 = 0 0 + 1 = 1 0 + 2 = 2 0 + 3 = 3 0 + 4 = 4 0 + 5 = 5
1 + 0 = 1 1 + 1 = 2 1 + 2 = 3 1 + 3 = 4 1 + 4 = 5
2 + 0 = 2 2 + 1 = 3 2 + 2 = 4 2 + 3 = 5
3 + 0 = 3 3 + 1 = 4 3 + 2 = 5
4 + 0 = 4 4 + 1 = 5
5 + 0 = 5
这道题虽然AC了，总感觉递归的姿势不正确= =
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
using namespace std;
int N;
void func(int num,int cnt)
{
if (num <= N)
{
if (num + cnt <= N)
{
printf("%-2d+%2d = %-2d ",num,cnt,num+cnt);
//cout << setiosflags(ios::left) << setw(2) << num << " + " << setiosflags(ios::right) << setw(2) << cnt << " = " << setiosflags(ios::left) << setw(2) << num+cnt << " ";
func(num,cnt+1);
}
cout << endl;
}
if (num > N)
exit(0);
func(num+1,0);
}
int main()
{
scanf("%d",&N);
func(0,0);
return 0;
}