D - Area解题报告

D - Area

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the following vertexes of the polygon until back to the initial vertex. For each step you may go North, West, South or East with step length of 1 unit, or go Northwest, Northeast, Southwest or Southeast with step length of square root of 2. 

For example, this is a legal polygon to be computed and its area is 2.5: 

Input

The first line of input is an integer t (1 <= t <= 20), the number of the test polygons. Each of the following lines contains a string composed of digits 1-9 describing how the polygon is formed by walking from the origin. Here 8, 2, 6 and 4 represent North, South, East and West, while 9, 7, 3 and 1 denote Northeast, Northwest, Southeast and Southwest respectively. Number 5 only appears at the end of the sequence indicating the stop of walking. You may assume that the input polygon is valid which means that the endpoint is always the start point and the sides of the polygon are not cross to each other.Each line may contain up to 1000000 digits.

Output

For each polygon, print its area on a single line.

Sample Input

4582567256244865

Sample Output

000.52

这个题目主要是涉及到了精度的问题，只能用__int64;

 

代码1：

#include<iostream>using namespace std;__int64 move[10][2]={{0,0},{1,-1},{1,0},{1,1},{0,-1},{0,0},{0,1},{-1,-1},{-1,0},{-1,1}};__int64 Getarea(__int64 x1,__int64 x2,__int64 y1,__int64 y2){return (x1*y2-x2*y1);}int main(){int n,p;char c;__int64 x1=0,x2=0,y1,y2,area=0;cin>>n;while(n--){while((c=cin.get())!='5'){p=c-'0';x2=x1+move[p][0];y2=y1+move[p][1];area+=Getarea(x1,x2,y1,y2);x1=x2;y1=y2;}if(area<0){area=(-1)*area;}x2=area/2;if(area!=x2*2){printf("%I64d.5\n",x2);}elseprintf("%I64d\n",x2);area=0;}return 0;}

代码2：

#include<iostream>using namespace std;struct Point{    __int64 x;    __int64 y;};int D[10][2]={{0,0},{-1,-1},{0,-1},{1,-1},{-1,0},{0,0},{1,0},{-1,1},{0,1},{1,1}};//求面积__int64 area(Point p1,Point p2){    return p1.x*p2.y-p2.x*p1.y;}int t;char ch[1000005];int main(){    int i,j;    int id;    int len;    __int64 Area=0;    scanf("%d",&t);    Point p0,p1;    for(i=0;i<t;++i)    {        p0.x=0;        p0.y=0;        Area=0;        scanf("%s",&ch);        len=strlen(ch);        for(j=0;j<len-1;++j)        {            id=ch[j]-'0';            p1.x=p0.x+D[id][0];            p1.y=p0.y+D[id][1];            Area+=area(p1,p0);            p0=p1;        }        if(Area<0)            Area=-Area;        __int64 res=Area/2;        if(res*2 == Area)            printf("%I64d\n",res);        else            printf("%I64d.5\n",res);    }    return 0;}