PAT (Advanced Level) 1060 Are They Equal
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题解
找出有效的字符串(t),第一个非零数字的位置(zero)和小数点的位置(point)。
若 s = "0012.104654",N = 4 则 t = "12104654",zero = 2,point = 4,所以答案为 0.1210*10^(point-zero)
若 s = "0000.00010465",N = 4 则 t = "10465",zero = 8,point = 4,所以答案为 0.1046*10^(point-zero+1)
注意 s = "0.0000000"的情况,若 N = 3,则答案为0.000*10^0
代码
#include<bits/stdc++.h>
using namespace std;
string process(int N,string s,int & k);
int main()
{
int i,N,k1,k2;
string str1,str2;
cin>>N>>str1>>str2;
str1=process(N,str1,k1);
str2=process(N,str2,k2);
if(str1==str2 && k1==k2) printf("YES %s*10^%d",str1.c_str(),k1);
else printf("NO %s*10^%d %s*10^%d",str1.c_str(),k1,str2.c_str(),k2);
system("pause");
return 0;
}
string process(int N,string s,int & k)
{
int i,zero,point;
string t;
zero=point=-1;
for(i=0;i<s.size();i++)
{
if(s[i]==‘.‘) point=i;
else if(s[i]==‘0‘ && zero==-1) continue;
else
{
if(zero==-1) zero=i;
t+=s[i];
}
}
if(zero==-1)
{
t.insert(0,N,‘0‘);
k=0;
return "0."+t;
}
else
{
if(point==-1) point=i;
if(t.size()<N) t.insert(t.size(),t.size()-N,‘0‘);
if(zero>point) k=point-zero+1;
else k=point-zero;
return "0."+t.substr(0,N);
}
}
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