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<$2^{31}$,0<=P<$2^{31}$)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 longusing 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;}