记录当前点和前一个点的坐标，算叉积，然后加入总面积之中
注意最后得到的面积有可能是负的，要取绝对值，还有答案有可能超过 int 范围，要用 long long
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-13 11:36:48
File Name: pku1654.cpp
Description:
************************************************************************/
#include <iostream>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{ for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) { for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
const int dx[10] = {0, -1, 0, 1, -1, 0, 1, -1, 0, 1};
const int dy[10] = {0, -1, -1, -1, 0, 0, 0, 1, 1, 1};
char s[1000010];
int main()
{
int ca;
for (scanf("%d", &ca); ca--;)
{
scanf("%s", s);
long long x = 0, y = 0, px = 0, py = 0, area = 0;
for (int i = 0; s[i] != '5'; i++)
{
x += dx[s[i] - '0'];
y += dy[s[i] - '0'];
area += x * py - y * px;
px = x;
py = y;
}
if (area < 0) area = -area;
if (area % 2) printf("%I64d.5\n", area / 2);
else printf("%I64d\n", area / 2);
}
return 0;
}