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using scanf is a better option since scanf is said to faster than cin !

and reading until EOF is a factor...while(scanf("%d",&N)!=EOF) can help

I got TLE with my code.... Can someone help???? #include<cmath> #include<iostream> #include<cstdio> using namespace std; int main() { int N, i, c = 0; while(1) { if(c != 0) cout << endl; cin >> N; c++; int res = 0; float dogx, dogy, gopx, gopy, escapex, escapey; float holex[N], holey[N]; cin >> gop...

I got TLE with my code.... Can someone help???? #include<cmath> #include<iostream> #include<cstdio> using namespace std; int main() { int N, i, c = 0; while(1) { if(c != 0) cout << endl; cin >> N; c++; int res = 0; float dogx, dogy, gopx, gopy, escapex, escapey; float holex[N], holey[N]; cin >> gopx...

Sorry i was kidding.

whats the logic behind this line? if(((dog_x>=gophar_x)&&(gophar_x>=hole_x[i]))||((dog_x<=gophar_x)&&(gophar_x<=hole_x[i]))||((dog_x-gophar_x==0)&&(((dog_y>=gophar_y)&&(gophar_y>=hole_y[i]))||((dog_y<=gophar_y)&&(gophar_y<=hole_y[i]))))) just comment this line...

Can some one help me. I got TLE..

oh.. thats the easy part.. you can do like this.. typedef struct Hole { double x, y; } Hole hole[MAXHOLES]; for (int i = 0; i < nholes; i++) { scanf("%lf %lf", &hole[i].x, &hole[i].y); } for (int i = 0; i < nholes; i++) { // if feasible with this hole then set ur flag and break.. }

NEWKID,
IF NOT BREAK THE LOOP
THEN IT PRINT THE LAST ESCAPABLE HOLE.
IF MORE THEN 1 ESCAPABLE HOLE THEN WHAT SHOULD BE ANSWER????

@sazzadcsedu you are breaking the " for(i=0;i<nhole;i++) " loop if you find a suitable hole for the gopher. this loop happens to do the scanning as well " scanf("%lf %lf",&h_x,&h_y); ". now tell me what will happen if you find a suitable hole before scanning all...

whats wrong with my code?? #include<stdio.h> #include<math.h> #include<stdlib.h> #define eps 1e-7 double h[1001][1001]; int main() { int nhole,i,flag; double x1,x2,y1,y2,dis_gof,dis_dog,h_x,h_y; // freopen("input.txt","r",stdin); //freopen("output.txt","w",std...

People, perhaps this may be the most useful tip on getting your 10310 program to AC. First of all, you can use doubles (no need to use long long and multiply input by 1000 blahblah). Secondly, when the dog & gopher reach the hole at same time, it's ...

shadow wrote:The problem hasnt precisely defined the answer when the dog and the gopher both reach a hole at the same time. I think the answer in this case is "can reach".

you're right..

The problem hasnt precisely defined the answer when the dog and the gopher both reach a hole at the same time. I think the answer in this case is "can reach".

my code is but is wrong !! #include <stdio.h> #include <iostream.h> #include <math.h> void main (void){ double dg,dp,hx,hy,gx,gy,px,py,rhx,rhy; int i,nhole; while (scanf("%d %lf %lf %lf %lf",&nhole,&gx,&gy,&px,&py)==5){ int k=0; for (i=1;i<=nhole;i++){ scanf("%lf %...

is this is a valid in c or c++
all are in double
r1=sqrt(pow(x1-h1,2.0)+pow(y1-o1,2.0));