目录
题目
Input Specification:
Output Specification:
Sample Input 1:
Sample Output 1:
Sample Input 2:
Sample Output 2:
思路
C++ 知识点UP
代码
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.d[1]...d[N]*10^k (d[1]>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思路
难度评级：⭐️⭐️
1. 先去小数点
比如123.456变成123456*10^3
2. 再整数保留n位（比如n为3时）
即123456*10^3变成123*10^3
3. 最后确定次方数
即123*10^3中123为格式化后的小数部分，次方数由3变为3+n=6，最终为0.123*10^6
C++ 知识点UP
1. vector判断是否相等
vector相等是指容器内值和顺序都一致
可以直接用vec1==vect2这种方式去判断
2. 指针类型变量所指地址改值（注意*优先级）
比如：指针类型变量power所指地址的值需要+1时，不可以直接*power++，而要对*power加括号，即(*power)++，因为*的优先级低于++的优先级
代码
#include <iostream>
#include <vector>
#include <cmath>using namespace std;int n;vector<char> formatFloat(string str,int* power) {vector<char> res;// 先去小数点（eg.将123.456变成123456*10^-3） bool isBegin=true;bool afterPot=false; (*power)=0;for(int i=0;i<str.size();i++) {// 舍弃掉开头连续的0 if(isBegin&&str[i]=='0') {// 小数点后的0在舍弃时，power要变小 if(afterPot) (*power)--;continue;}// 遇到小数点时 if(str[i]=='.') {afterPot=true;}else {// 遇到数字时res.push_back(str[i]);if(afterPot) (*power)--;isBegin=false; } }// 这个数就是0时 if(isBegin) {res.push_back('0');*power=0;} // 计算整数部分只保留n位时的情况（eg.123456*10^-3变成123*10^-3） int diff=res.size()-n;if(!isBegin) (*power)+=diff; for(int i=0;i<abs(diff);i++) {if(diff>0) res.pop_back();else if(diff<0) res.push_back('0');}// res记录小数部分，power记录题目要求的format形式中的次方数if(!isBegin) (*power)+=n; return res;
}void printFormat(vector<char> vec, int power) {cout<<" 0.";for(int i=0;i<n;i++) {cout<<vec[i];}cout<<"*10^"<<power;
}int main(int argc, char** argv) {string a,b;cin>>n>>a>>b;int powerA,powerB;vector<char> aFormat=formatFloat(a,&powerA);vector<char> bFormat=formatFloat(b,&powerB);// 判断a b是否相等if(powerA==powerB&&aFormat==bFormat) {cout<<"YES";printFormat(aFormat,powerA); }else {cout<<"NO";printFormat(aFormat,powerA); printFormat(bFormat,powerB); }return 0;
}
