uva 846

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 = 0 ; i < n ; i++ ){
            scanf( "%d%d", &x, &y );

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

            step = (int)(sqrt((double)dis)+ERROR);
            if( step * step == dis ) step = step * 2 - 1;
            else if( step * step + step < dis ) step = step * 2 + 1;
            else step = step * 2;

            printf( "%d
", step );
        }
    }
    return 0;
}

另：

#include <iostream>
using namespace std;

int main()
{
    int x, y;
    int testCases;
    int min_steps = 0;
    cin >> testCases;
    while(testCases --)
    {
        cin >> x >> y;
        int difference = y - x;
        min_steps = 0;

        if(difference != 0)
        {
            int sumOfSteps = 0;
            int z = 2; //divided by 2, it represents the size if the next step

            while(difference > sumOfSteps)
            {
                sumOfSteps += (z / 2); // next step
                min_steps ++;
                z++;
            }
        }
        cout << min_steps << endl;
    }
    return 0;
}

#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 == 0)
            puts("0");
        else
        {
            n = (int)sqrt(diff);
            diff -= n * n;
            if (diff == 0)
                printf("%d
", (n << 1) - 1);
            else if (diff <= n)
                printf("%d
", n << 1);
            else
                printf("%d
", (n << 1) + 1);
        }
    }
    return 0;
}