1060. Are They Equal (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1060

题目：

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*10⁵ with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10¹⁰⁰, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line "YES" if the two numbers are treated equal, and then the number in the standard form "0.d₁...d_N*10^k" (d₁>0 unless the number is 0); or "NO" if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

分析：

0.abc...（a!=0）型科学计数法表示
此题因为数据超长需用字符串进行处理，有一些细节需要注意
1. 整数部分之前和小数部分之后可能都有多余的0
2.真值为0的数的指数都计作0
3.类似0.0008之类整数为0且小数部分有前导0的数要转化为0.8来考虑并相应修改指数
4.转化后的数可能后面要补0

AC代码：

#include<stdio.h>
#include<string>
using namespace std;
int getE(char * p){ //得到指数
 int e = 0;
 
 if (p[0] != '0'){
  for (int i = 0; p[i] != 0 && p[i] != '.'; i++)
   e++;
 }
 else if(p[1] == '.'){
  for (int i = 2; p[i + 1] != 0 && p[i] == '0'; i++){
   e--;
  }
 }
 else{
  int j = 0;
  while (p[j] == '0')j++;
  for (; p[j] != 0 && p[j] != '.'; j++)
   e++;
 }
 return e;
}
string getB(char *p, int n,int e){ //得到基数
 string ret = "";
 int idx = 0;
 if (e > 0){
  int i = 0;
  while (p[i] == '0')i++; //整数的话把前面多余的0去掉
  for (; p[i] != 0 && idx <n; i++){
   if (p[i] == '.')continue;
   ret += p[i];
   idx++;
  }
 }
 else {
  int i = 0;
  if (p[i + 1] == 0){
   ret = "0";
   while (ret.size() < n){
    ret += '0';
   }
   return ret;
  }
  while (p[i] != '.')i++;
  ++i;
  while (p[i] == '0')i++;  //0.000x的问题
  for (; p[i] != 0 && idx < n; i++){
   ret += p[i];
   idx++;
  }
 }
 while (ret.size() < n){
  ret += '0';  //不足n部分需要在后面添加0
 }
 return ret;
}
int main(){
 freopen("F://Temp/input.txt", "r", stdin);
 int N;
 char A[101];
 char B[101];
 scanf("%d %s %s", &N, A, B);
 string A2;
 string B2;
 int Ae = getE(A);
 int Be = getE(B);
 A2 = getB(A, N,Ae);
 B2 = getB(B, N,Be);
 if (N == 0){
  printf("YES 0.*10^0\n");
 }
 else if (Ae == Be && A2 == B2){
  printf("YES 0.%s*10^%d\n", A2.c_str(), Ae);
 }
 else{
  printf("NO 0.%s*10^%d 0.%s*10^%d\n", A2.c_str(), Ae, B2.c_str(), Be);
 }
 return 0;
}

截图：

在搞定了0.000x的问题

搞定了0的0.000*10^0的格式后

在解决了整数前面有0需要去除影响的情况后

——Apie陈小旭