HDU---1030-Delta-wave (简单数学)

Delta-wave

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 2

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Problem Description

A triangle field is numbered with successive integers in the way shown on the picture below.



The traveller needs to go from the cell with number M to the cell with number N. The traveller is able to enter the cell through cell edges only, he can not travel from cell to cell through vertices. The number of edges the traveller passes makes the length of the traveller's route.

Write the program to determine the length of the shortest route connecting cells with numbers N and M.

Input

Input contains two integer numbers M and N in the range from 1 to 1000000000 separated with space(s).

Output

Output should contain the length of the shortest route.

Sample Input

6 12 

Sample Output

3

Source

Ural Collegiate Programming Contest 1998

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<pre name="code" class="cpp">#include <iostream>#include <cmath>#include <cstdio>#include <iomanip>using namespace std;int main(){    int a,b;  // 10亿    while(cin>>a>>b)    {        if(a>b)swap(a,b);        int sum;        int x=pow((double)(a-1),0.5)+1;  // 顶部当前行        int y=pow((double)(b-1),0.5)+1;  // 底部当前行        if(x==y)        {            cout<<b-a<<endl;            continue;        }        int ylen = y-x;            //行差        int xx=a-(x-1)*(x-1);      //当前坐标        int yy=b-(y-1)*(y-1);      //当前坐标        int sum1,sum2;        if((xx%2==1 && yy<=xx+2*ylen && yy>=xx) || (xx%2==0 && yy<=xx+2*ylen+1 && yy>=xx-1))  //内部覆盖判断        {            sum=2*ylen;            if(yy%2==0)--sum;      // 底部倒三角            if(xx%2==1)--sum;      // 顶部正三角        }        else                       // 外部        {            if(xx%2==0)            //顶部倒三角            {                if(yy<xx-1)        //覆盖区域左边                    sum=xx-1-yy+2*ylen;                else                    sum=yy-xx-1;   //右边            }            else                   //顶部正三角            {                if(yy<xx)          //覆盖区域左边                    sum=xx-yy+2*ylen-1;                else                    sum=yy-xx-1;   //右边            }        }        cout<<sum+1<<endl;    }    return 0;}