PAT A1060 Are They Equal
tag: [string]
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
- 生词
word | meaning |
---|---|
with simple chopping | 简单的切割 |
non-negetive | 非负 |
assume | v.假设 |
without rounding | 没有四舍五入 |
- 思路:
用string保存浮点数,先把多余的前导零去掉,去掉后剩余部分,根据小数点位置分为3种情况:
1)【原】0000.12345 -> .12345 或 .0012345
2)【原】000012.345 -> 12.345
3)【原】000012345 -> 12345 //相当于小数点在最末尾
分别处理,首先计算出指数e:
1)中e为.后面第一个有效数字(非0)的位置减1(如.123
为e=1-1=0
, .00123
为e=3-1=2
),可以通过不断从前面截断字符串的方式实现;2) 和 3)e就是小数点的位置(如12.345
为e=2
)
切割出有效位(获得子串),分为2种情况:
1)有效位数够 -> 直接截取前N(去掉.
)
2)有效位不足 -> 末尾补0(用循环不断s += '0'
)
最后,比较,格式化输出即可;
- TIPS:
注意不要落掉 0000 0000.00这种情况,如果全0 -> 经过上面对string的处理后,s为空,对应的字串就是精度要求个数的0,对应的指数e为0
- code1:
#include <string>
#include <iostream>
using namespace std;
string deal(string s, int &e, int N){
while(s.length() > 0 && s[0] == '0'){
s.erase(s.begin());
}
if(s[0] == '.'){
s.erase(s.begin());
while(s.length() > 0 && s[0] == '0'){
s.erase(s.begin());
e--;
}
}
else{
if(s.find(".") != string::npos){
e = s.find(".");
}
else{
int n = 0;
for(string::iterator it = s.begin(); it != s.end(); it++){
n++;
}
e = n;
}
}
if(s.length() == 0){
e = 0;
string sub1;
for(int i = 0; i < N; ++i){
sub1 += "0";
}
return sub1;
}
else{
if(s.find(".") != string::npos){
s.erase(s.begin() + s.find("."));
}
string sub = s.substr(0, N);
return sub;
}
}
void Output(string s1, string s2, int e1, int e2){
if(s1 == s2 && e1 == e2){
cout << "YES 0." << s1 << "*10^"
<< e1;
}
else{
cout << "NO 0." << s1 << "*10^"
<< e1 << " " << "0." << s2
<< "*10^" << e2;
}
}
int main(){
int N;
string s1, s2;
int e1 = 0, e2 = 0;
cin >> N >> s1 >> s2;
string sub1 = deal(s1, e1, N);
string sub2 = deal(s2, e2, N);
Output(sub1, sub2, e1, e2);
return 0;
}
- code2: 算法笔记
#include <string>
#include <iostream>
using namespace std;
int N, e1, e2;;
void deal(string &s, int &e){
while(s.length() > 0 && s[0] == '0'){
//Q1:len > 0 需要加吗?
//去掉前导零
s.erase(s.begin());
}
if(s[0] == '.'){
//如果是" .0012345 "类型的
s.erase(s.begin());
while(s[0] == '0'){
s.erase(s.begin());
e--;
}
}
else{
//如果是" 12345"或" 123.45"类型的
for(int i = 0; i < s.length(); ++i){
if(s[i] == '.'){
//如果找到" ." 删除"." 跳出循环
s.erase(s.begin() + i);
break;
}
e++;
}
}
if(s.length() == 0){
//如果全是由0组成,如"000.00"或"0000"
e = 0;
}
//经过处理后的string都变成了12345的形式
//下面获得长度N的子串,如果长度不足,+='0'
int len = s.length();
if(s.length() < N){
for(int i = 0; i < N-len; ++i){
s += '0';
}
}
else{
s.erase(N, s.length());
//或s.erase(s.begin()+N, s.end());
}
}
void equal(string s1, string s2){
if(s1 == s2 && e1 == e2){
cout << "YES 0." << s1 << "*10^" << e1;
}
else{
cout << "NO 0." << s1 << "*10^" << e1 <<
" 0." << s2 << "*10^" << e2;
}
}
int main(){
string s1, s2;
cin >> N >> s1 >> s2;
deal(s1, e1); deal(s2, e2);
equal(s1, s2);
return 0;
}
- 思路3:统一化:全转化为处理整数,初始 e=0
step 1:无论小数点在哪,先把小数点移到最末尾(转化为整数),如 0.00123 = 000123 * 10^-5
、 1.2300 = 12300 * 10 ^ -4
…, 即 小数点左移:e = 小数点到末尾的距离 = point_pos - s.size()
step 2: 问题转化为保留一个整数有效位的问题:000123 * 10 ^-5 = 0.123 * 10 ^ -2
12300 * 10 ^ -4 = 0.123 * 10 ^ 1
…
即:去掉前导零0后!把小数点右移到最前面:e += s.size()
【注意】去掉前导零后字符串可能为空,说明输入为: “0”,应令 e = 0;
- 测试样例:
样例 2 : 3 122 122.00
output: YES 0.122*10^3
样例 3: 3 0.12 0.012
output: NO 0.120*10^0 0.120*10^-1
样例 4: 5 00643.0008 0643.0
output: YES 0.64300*10^5
样例 6 : 6 000000000 000000.00
output: YES 0.000000*10^0
- T2 code:
#include <bits/stdc++.h>
#include <string>
using namespace std;
void Change(string & s, int & e, int n){
int pos = s.find(".");
if(pos != string::npos){
e = (pos + 1) - s.size();
s.erase(pos, 1);
}
while(s[0] == '0') s.erase(0, 1);
if(s.size() == 0) e = 0;
else e += s.size();
// e += s.size(); //Wrong 1:未考虑0: 4 0000 0000.00
if(s.size() < n){
string tmp(n - s.size(), '0');
s += tmp;
}else{
s = s.substr(0, n);
}
}
int main(){
int n, e1 = 0, e2 = 0;
string s1, s2;
cin >> n >> s1 >> s2;
Change(s1, e1, n);
Change(s2, e2, n);
if(s1 == s2 && e1 == e2){
printf("YES 0.%s*10^%d", s1.c_str(), e1);
}else{
printf("NO 0.%s*10^%d 0.%s*10^%d", s1.c_str(), e1, s2.c_str(), e2);
}
return 0;
}
- T3 code:
#include <bits/stdc++.h>
using namespace std;
string GetE(string s, int n)
{
int pos_point = s.size(), pos_value = -1;
bool lock = true;
for(int i = 0; i < s.size(); ++i)
{
if(s[i] == '.') pos_point = i;
if(lock && '0' < s[i] && s[i] <= '9')
{
pos_value = i;
lock = false;
}
}
int e, cnt = 0;
string ans = "0.", ans_suffix = "*10^";
if(pos_value == -1) //无效输入: 0000000 00000.0000
{
e = 0;
}else
{
e = pos_point - pos_value;
for(int i = pos_value; cnt < n && i < s.size(); ++i)
{
if(s[i] != '.')
{
ans.push_back(s[i]);
cnt++;
}
}
}
while(cnt < n)
{
ans.push_back('0');
cnt++;
}
e = e < 0 ? e + 1 : e;
return ans + ans_suffix + to_string(e);
}
int main()
{
int n;
scanf("%d", &n);
string num1, num2;
cin >> num1 >> num2;
string ans1 = GetE(num1, n);
string ans2 = GetE(num2, n);
if(ans1 == ans2)
{
printf("YES %s", ans1.c_str());
}else
{
printf("NO %s %s", ans1.c_str(), ans2.c_str());
}
return 0;
}
- T4 code:
#include <bits/stdc++.h>
using namespace std;
void Change(string & s, int & e)
{
auto it = s.find(".");
if(it != s.npos)
{
e = -(s.size() - 1 - it);
s.erase(it, 1);
}
while(s[0] == '0') s.erase(0, 1);
if(s.size() == 0)
{
e = 0;
}else
{
e += s.size();
}
}
void Judge(string s1, string s2, int e1,int e2, int n)
{
string sub1 = s1.substr(0, n), sub2 = s2.substr(0, n);
while(sub1.size() < n) sub1 += '0';
while(sub2.size() < n) sub2 += '0';
if(sub1 == sub2 && e1 == e2)
{
printf("YES 0.%s*10^%d", sub1.c_str(), e1);
}else
{
printf("NO 0.%s*10^%d 0.%s*10^%d", sub1.c_str(), e1, sub2.c_str(), e2);
}
}
int main()
{
int n;
scanf("%d", &n);
string n1, n2;
int e1 = 0, e2 = 0;
cin >> n1 >> n2;
Change(n1, e1);
Change(n2, e2);
Judge(n1, n2, e1, e2, n);
return 0;
}