uva 846 - Steps

找出步數與距離的關係即可得解。

• 0步最多能抵達的距離是0
• 1步最多能抵達的距離是1(1)
• 2步最多能抵達的距離是2(1 1)
• 3步最多能抵達的距離是4(1 2 1)
• 4步最多能抵達的距離是6(1 2 2 1)
• 5步最多能抵達的距離是9(1 2 3 2 1)
• 6步最多能抵達的距離是12(1 2 3 3 2 1)
• ……以此類推

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #define ERROR 1e-10
 using namespace std;
 int main(){
     int n, x, y;
     int dis, step;
     while( scanf( "%d", &n ) != EOF ){
         for( int i =  ; i < n ; i++ ){
             scanf( "%d%d", &x, &y );

             dis = y-x;
             if( dis ==  ){
                 printf( "0\n" );
                 continue;
             }

             step = (int)(sqrt((double)dis)+ERROR);
             if( step * step == dis ) step = step *  - ;
             else if( step * step + step < dis ) step = step *  + ;
             else step = step * ;

             printf( "%d\n", step );
         }
     }
     return ;
 }

另：

 #include <iostream>
 using namespace std;

 int main()
 {
     int x, y;
     int testCases;
     int min_steps = ;
     cin >> testCases;
     while(testCases --)
     {
         cin >> x >> y;
         int difference = y - x;
         min_steps = ;

         if(difference != )
         {
             int sumOfSteps = ;
             int z = ; //divided by 2, it represents the size if the next step

             while(difference > sumOfSteps)
             {
                 sumOfSteps += (z / ); // next step
                 min_steps ++;
                 z++;
             }
         }
         cout << min_steps << endl;
     }
     return ;
 }

 #include<cstdio>
 #include<cmath>  

 int main(void)
 {
     int t, x, y, diff, n;
     scanf("%d", &t);
     while (t--)
     {
         scanf("%d%d", &x, &y);
         diff = y - x;
         if (diff == )
             puts("");
         else
         {
             n = (int)sqrt(diff);
             diff -= n * n;
             if (diff == )
                 printf("%d\n", (n << ) - );
             else if (diff <= n)
                 printf("%d\n", n << );
             else
                 printf("%d\n", (n << ) + );
         }
     }
     return ;
 }