Codeforces #3 A Shortest path of the king

来源：互联网发布：高中生很敏感知乎编辑：程序博客网时间：2024/05/30 23:25

题目链接 http://codeforces.com/problemset/problem/3/A
走过的格的值相乘后后缀的0最少，统计路径中中 2和5最少的数就行，有值0的话看另外判断，可能有路径后缀的0没有。
下面ac代码

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <string>#include <map>using namespace std;typedef long long lld;/*方向 R U L D RU RD LU LD       0 1 2 3 4  5  6  7*/int main(){    char x1,x2;    int  y1,y2;    int dir[8]={0};    while(cin>>x1>>y1>>x2>>y2)    {        memset(dir,0,sizeof(0));        int xcha=x2-x1;        int ycha=y2-y1;        if(xcha==0&&ycha==0)           cout<<0;        else if(xcha>=0&&ycha>=0)        {            int flag=0;            while(xcha>0&&ycha>0)            {                ycha--;                xcha--;                dir[4]++;                flag++;            }            while(xcha>0)            {                xcha--;                dir[0]++;                flag++;            }            while(ycha>0)            {                ycha--;                dir[1]++;                flag++;            }            cout<<flag<<endl;        }        else if(xcha<=0&&ycha<=0)        {            int flag=0;            while(xcha<0&&ycha<0)            {                ycha++;                xcha++;                dir[7]++;                flag++;            }            while(xcha<0)            {                xcha++;                dir[2]++;                flag++;            }            while(ycha<0)            {                ycha++;                dir[3]++;                flag++;            }            cout<<flag<<endl;        }        else if(xcha<=0&&ycha>=0)        {            int flag=0;            while(xcha<0&&ycha>0)            {                ycha--;                xcha++;                dir[6]++;                flag++;            }            while(xcha<0)            {                xcha++;                dir[2]++;                flag++;            }            while(ycha>0)            {                ycha--;                dir[1]++;                flag++;            }            cout<<flag<<endl;        }        else if(xcha>=0&&ycha<=0)        {            int flag=0;            while(xcha>0&&ycha<0)            {                ycha++;                xcha--;                dir[5]++;                flag++;            }            while(xcha>0)            {                xcha--;                dir[0]++;                flag++;            }            while(ycha<0)            {                ycha++;                dir[3]++;                flag++;            }            cout<<flag<<endl;        }        while(dir[0]--)            cout<<"R"<<endl;        while(dir[1]--)            cout<<"U"<<endl;        while(dir[2]--)            cout<<"L"<<endl;;        while(dir[3]--)            cout<<"D"<<endl;;        while(dir[4]--)            cout<<"RU"<<endl;;        while(dir[5]--)            cout<<"RD"<<endl;;        while(dir[6]--)            cout<<"LU"<<endl;;        while(dir[7]--)            cout<<"LD"<<endl;;        cout<<endl;/*方向 R U L D RU RD LU LD       0 1 2 3 4  5  6  7*/    }    return 0;}

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