C ++比较器保证和内部运作

Question

Solving this problem , I found that

if(!(p<arr[i]))

and

if(p>=arr[i])

May have different results( p is a long long and arr[i] is a double ), since the first solution is accepted and the second is not. Why?

Complete code for context:

#include <bits/stdc++.h>
#define EPS (1e-5)
using namespace std;
typedef long long ll;
double arr[10005];
int main(){
    ll D,p;
    string s;
    while(getline(cin,s)){
        stringstream ss(s);
        ss>>D>>p;
        int n=0;
        while(ss>>arr[n]) ++n;
        ll dmin=D+1;
        if(n<D+1){
            double a=arr[n-4];
            double b=arr[n-3];
            double c=arr[n-2];
            double d=arr[n-1];
            double den=a*c-b*b;
            double s=(c*c-b*d)/den;
            double t=(a*d-b*c)/den;
            for(int i=n;i<=D;++i)
                arr[i]=s*arr[i-2]+t*arr[i-1];
        }
        for(int i=0;i<=D;++i){
            if(!(p<arr[i]))
                dmin=min(dmin,D-i);
            else
                break;
        }
        if(dmin==0) cout<<"The spider may fall!"<<endl;
        else if(dmin==D+1) cout<<"The spider is going to fall!"<<endl;
        else cout<<dmin<<endl;
    }
}

Answer 1

If !(a<b) and (a>=b) are different then the most likely explanation is that one of them is a NaN. Any comparison involving a NaN results in false , so if b is a NaN, then both (a<b) and (a>=b) will be false, and !(a<b) will be true.

Looking at your code, first:

double den=a*c-b*b;
double s=(c*c-b*d)/den;
double t=(a*d-b*c)/den;

There is no obvious reason why den cannot be zero, in which case both s and t will be infinities. Then, you compute:

for(int i=n;i<=D;++i)
  arr[i]=s*arr[i-2]+t*arr[i-1];

If s and t are inifinities, then there is ample room for the sum which produces arr[i] to be a NaN. The sum of two infinities of the same sign is the infinity itself, but the sum of two infinities of opposite signs is a NaN. Also, the product of 0 and infinity is a NaN.

And once a NaN gets into arr , it will propagate, since each element depends on the previous one.

So you probably need to do something if den is 0.