Codeforces 11B Jumping Jack (thinking)

Test instructions: from 0 on the axis can go to left and right, the distance of step I is I, to reach the minimum number of steps x (X<10^9);

Idea: Consider always to the right, if the exact arrival is the minimum number of steps;

If the distance is greater than X, and (distance-X) =n,n is an even number, it should be moved to the left N;

To move the distance N to the left, simply walk to the left of step N/2;

if (Journey-X) is odd, go straight to the even number of cases;

#include <cstdio>#include<cstring>#include<algorithm>using namespacestd;intn,m,sum;intMain () {inti,j,k;  while(SCANF ("%d", &n)! =EOF) {        if(n<0) n=-N; //printf ("%d\n", n);sum=0;  for(i=0;; i++) {sum+=i; if(sum==n) Break; if(sum>n&& (sum-n)%2==0) Break; } printf ("%d\n", i); }    return 0;}

Codeforces 11B Jumping Jack (thinking)