题意：
画一个a*b*c的三维长方体。
思路：
把每一条线的起始坐标和终点坐标计算出来，画上去。
代码：
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <sstream>
#define pb push_back
#define X first
#define Y second
#define lch (o<<1)
#define rch (o<<1|1)
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pii pair<int,int>
#define qclear(a) while(!a.empty())a.pop();
#define lowbit(x) (x&-x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define mst(a,b) memset(a,b,sizeof(a))
#define cout3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl
#define cout2(x,y) cout<<x<<" "<<y<<endl
#define cout1(x) cout<<x<<endl
#define IOS std::ios::sync_with_stdio(false)
#define SRAND srand((unsigned int)(time(0)))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
using namespace std;
const double PI=acos(-1.0);
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
const double eps=1e-8;
const int maxn=100005;
const int maxm=10005;
int a,b,c;
char maps[100][100];
void out() {
for(int i=1; i<=2*b+2*c+1; i++) {
for(int j=1; j<=2*a+2*b+1; j++) {
printf("%c",maps[i][j]);
}
printf("\n");
}
}
void solve() {
int t;
sd(t);
while(t--) {
sddd(a,b,c);
mst(maps,'.');
int n=2*b+2*a+1;
for(int i=1; i<=2*b-1; i+=2) {
int cnt=a+1;
for(int j=2*b+2-i; cnt; j+=2) {
maps[i][j]='+';
cnt--;
}
cnt=a;
for(int j=2*b+3-i; cnt; j+=2) {
maps[i][j]='-';
cnt--;
}
}
int cnt=c+1;
for(int i=2*b+1; cnt; i+=2) {
int cnt1=a+1;
for(int j=1; cnt1; j+=2) {
maps[i][j]='+';
cnt1--;
}
int cnt2=a;
for(int j=2; cnt2; j+=2) {
maps[i][j]='-';
cnt2--;
}
cnt--;
}
cnt=b;
for(int i=2; cnt; i+=2) {
int cnt1=a+1;
for(int j=2*b+2-i; cnt1; j+=2) {
maps[i][j]='/';
cnt1--;
}
cnt--;
}
cnt=c;
for(int i=2*b+2; cnt; i+=2) {
int cnt1=a+1;
for(int j=1; cnt1; j+=2) {
maps[i][j]='|';
cnt1--;
}
cnt--;
}
cnt=b;
for(int j=2*b+2*a+1; cnt; j-=2) {
int cnt1=c+1;
for(int i=2*c+1+(2*a+2*b+1-j); cnt1; i-=2) {
maps[i][j]='+';
cnt1--;
}
cnt1=c;
for(int i=2*c+(2*a+2*b+1-j); cnt1; i-=2) {
maps[i][j]='|';
cnt1--;
}
cnt--;
}
cnt=b;
for(int j=2*b+2*a; cnt; j-=2) {
int cnt1=c+1;
for(int i=2*c+1+(2*a+2*b+1-j); cnt1; i-=2) {
maps[i][j]='/';
cnt1--;
}
cnt--;
}
out();
}
return ;
}
int main() {
#ifdef LOCAL
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#else
// freopen("","r",stdin);
// freopen("","w",stdout);
#endif
solve();
return 0;
}