玲珑杯 J -- Just for Fun

J -- Just for Fun

Time Limit：1s Memory Limit：64MByte

Submissions：970Solved：75

DESCRIPTION

P=NP?

INPUT

First line contains a single integer T which denotes the number of test cases. For each test case, there is two numbers which denote N and P separated by a space.(0<=N<231231,0<=P<231231)The
input data is no large than 1M.

OUTPUT

For each case, output the "Y" or "N" in a single line.

SAMPLE INPUT

21 123123 0

SAMPLE OUTPUT

YY

SOLUTION

玲珑杯”ACM比赛 Round #5

本是水题，奈何坑死一群人，其实看仔细点就能发现坑点，因为题目没有说是整数，而且长度不定，有可能是0000.0000000或者0000001.000000000000，所以有人用Long long和long double存，其实都是存不下的。只能用大数模拟，所以只要判断前面的数为1或后面的数为0就AC了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define mem(p,k) memset(p,k,sizeof(p))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define inf 0x5fffffff
#define LL long long
using namespace std;
char n[5000000],p[5000000];
int main(){
int t;
cin>>t;
while(t--){
scanf("%s%s",&n,&p);//cout<<n<<p<<endl;
int cur=0,cur2=0;
for(int i=0;i<strlen(n);i++){

if(n[i]=='.'){
for(int j=i+1;j<strlen(n);j++){
if(n[j]!=48){
cur=1;break;
}
}
break;
}
else if(n[i]=='1'){
cur2=1;
if(i!=strlen(n)-1&&n[i+1]!='.'){

cur=1;break;
}
}
else if(n[i]!=48){//cout<<"asd"<<endl;
cur=1;break;
}

}//cout<<cur<<endl;
if(!cur&&cur2==1){
cout<<"Y"<<endl;continue;
}
cur=1;
for(int i=0;i<strlen(p);i++){
if(p[i]!='.'&&p[i]!=48){
cur=0;break;
}//cout<<"asd"<<endl;
}
if(!cur)cout<<"N"<<endl;
else cout<<"Y"<<endl;

}
return 0;
}