杭电的ACM 题

杭电的ACM 题 求助
http://acm.hdu.edu.cn/showproblem.php?pid=1030  

这道题 我卡住一天了 ，样例和自己想的测试样例都可以过

题意是 一个三角形 里面很多个块 ，从1开始编号，那么给出 两个值a b 问你 从a走到 b 要多久 

只能从 边走 ，不能从顶点跨过 

C/C++ code

# include<stdio.h># include<math.h>int main(){    //freopen("in.txt","r",stdin);freopen("myout.txt","w",stdout);    long a,b;    long i = 0,j=0;__int64 sa = 0,sb=0;    long step = 0,temp = 0;    while(scanf("%ld%ld",&a,&b)!=EOF)    {        i = (int)sqrt(a) ;j = (int)sqrt(b) ;        sa = a - i*i; sb = b - j*j;        i ++,j ++; // 计算出 行数        if(sa == 0) //当a  刚好是 平方数 的时候 行数要减一        {            i --;sa = 2*i - 1;        }        if(sb == 0)        {    j--;sb = 2*j -1; }        step = (j-i)*2;  // 到达j 行所要的步数        if(sa >= sb) // 当啊 的起始位置 大于 B 时            step += sa - sb;        else if(sa < sb)        {            if((sb-sa) < step) //这里考虑特殊情况            {                if(sa%2==1 && sb%2 ==0)                    step --;                else if(sa%2==0 && sb%2==1)                    step ++;            }            else            {                temp = sb - sa - step;                step += temp;            }        }        printf("%ld\n",step);    }    return 0;}

C/C++ code

// 1234.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <iostream>#include <cmath>using namespace std;int _tmain(int argc, _TCHAR* argv[]){    int n;    int m;    while(1){        cin>>n>>m;        if(m < n){//确保m不小于n            unsigned int iTemp = n;            n = m;            m = iTemp;        }        int iRow_n,iRow_m,iColumn_n,iColumn_m;        iRow_n = (int) sqrt(double (n));        iColumn_n = n - iRow_n * iRow_n;        if(iColumn_n){            -- iColumn_n;        }        else{            -- iRow_n;            iColumn_n = n - iRow_n * iRow_n - 1;        }        iRow_m = (int) sqrt(double (m));        iColumn_m = m - iRow_m * iRow_m;        if(iColumn_m){            -- iColumn_m;        }        else{            -- iRow_m;            iColumn_m = m - iRow_m * iRow_m - 1;        }        int iColumn_n_Left = (iRow_n + 1) * (iRow_n + 1) - iRow_n * iRow_n - 1 - iColumn_n;        int iColumn_m_Max = (iRow_m + 1) * (iRow_m + 1) - iRow_m * iRow_m;        int iColumn_m_Left = iColumn_m_Max - 1 - iColumn_m;        int iColumn_n_Right = iColumn_n;        int iColumn_m_Right = iColumn_m;        int iStep;        if(iRow_m == iRow_n){            iStep = abs(iColumn_m - iColumn_n);        }        else{            if(iColumn_m < iColumn_n){                iStep = 2 * (iRow_m - iRow_n) + iColumn_n - iColumn_m;            }            else{                if(iColumn_m < iColumn_n + iRow_n + iRow_n){                    iStep = 2 * (iRow_m - iRow_n - 1);                    int iOffset = iColumn_m % 2 * 2 + iColumn_n % 2;                    switch(iOffset){                        case 0:                            iStep += 2;                            break;                        case 1:                            iStep += 3;                            break;                        case 2:                            iStep += 1;                            break;                        case 3:                            iStep += 2;                            break;                        default :                            break;                    }                }                else{                    iStep = 2 * (iRow_m - iRow_n) + iColumn_m - (iColumn_n + iRow_n + iRow_n);                }            }        }        //cout<<"Row"<<iRow_n<<" "<<iRow_m<<endl;        //cout<<"Column"<<iColumn_n<<" "<<iColumn_m<<endl;        //cout<<"iStep"<<iStep<<endl;        cout<<iStep<<endl;    }    return 0;} 
 
[解决办法]
此题考点:DP(动态规划)
[解决办法]
C/C++ code
#include<iostream>#include<cmath>using namespace std;int main(){    int slayer,elayer,cs,start,end;    int sindex,eindex,width,height;    while (cin>>start>>end)       {        if (start>end)        {            int temp = start;            start = end;            end = temp;           }        slayer = (int)ceil(sqrt((double)start)); //开根号取最小整数（ceil()函数）找出开始行        elayer = (int)ceil(sqrt((double)end));                     //注意：求平方的时候如果用Pow()来求，可能会错        sindex = start - (slayer-1)*(slayer-1); //找出此数在当前行的位置。           eindex = end - (elayer-1)*(elayer-1);        height = elayer - slayer;  //两层之间的垂直距离        //用width和height比较可以确定end是否在start的2*height的辐射范围内。        width = abs(eindex-sindex-height);         cs = 2*height;        if (width>=height)        {            cs += width - height;  //cs += abs(eindex-sindex-height)-height;           }        else        {             if(elayer%2==0&&end%2!=0||elayer%2!=0&&end%2==0)                cs--;  //判断elayer层的上层还是下层三角形里，从而是否减1            if(slayer%2==0&&start%2!=0||slayer%2!=0&&start%2==0)                cs++; //判断elayer层的上层还是下层三角形里，从而是否加1        }        cout<<cs<<endl;    }    return 0;}