1 Show Code - Run ID 1166912
2
3 Submit Time: 2012-02-18 01:33:04 Language: GNU C Result: Accepted
4 Pid: 3124 Time: 0.00 sec. Memory: 852 K. Code Length: 0.6 K.
5 # include <stdio.h>
6 # define cross(x1,y1,x2,y2) (x1)*(y2)-(x2)*(y1)
7 # define get_aera(x0,y0,x1,y1,x2,y2) (cross((x1)-(x0),(y1)-(y0),(x2)-(x0),(y2)-(y0)))
8 int main()
9 {
10 int n;
11 while(scanf("%d",&n)!=EOF&&n)
12 {
13 int i;
14
15 double x[3],y[3],aera=0;
16 scanf("%lf%lf",&x[2],&y[2]);
17 for(i=1;i<n;i++)
18 {
19 scanf("%lf%lf",&x[i%2],&y[i%2]);
20 if(i>1) aera+=get_aera(x[2],y[2],x[(i+1)%2],y[(i+1)%2],x[i%2],y[i%2]);
21 }
22 aera*=0.5;
23 if(aera<0) aera=-aera;
24 printf("%.0f\n",aera+1e-8);
25
26 }
27 return 0;
28 }
2
3 Submit Time: 2012-02-18 01:33:04 Language: GNU C Result: Accepted
4 Pid: 3124 Time: 0.00 sec. Memory: 852 K. Code Length: 0.6 K.
5 # include <stdio.h>
6 # define cross(x1,y1,x2,y2) (x1)*(y2)-(x2)*(y1)
7 # define get_aera(x0,y0,x1,y1,x2,y2) (cross((x1)-(x0),(y1)-(y0),(x2)-(x0),(y2)-(y0)))
8 int main()
9 {
10 int n;
11 while(scanf("%d",&n)!=EOF&&n)
12 {
13 int i;
14
15 double x[3],y[3],aera=0;
16 scanf("%lf%lf",&x[2],&y[2]);
17 for(i=1;i<n;i++)
18 {
19 scanf("%lf%lf",&x[i%2],&y[i%2]);
20 if(i>1) aera+=get_aera(x[2],y[2],x[(i+1)%2],y[(i+1)%2],x[i%2],y[i%2]);
21 }
22 aera*=0.5;
23 if(aera<0) aera=-aera;
24 printf("%.0f\n",aera+1e-8);
25
26 }
27 return 0;
28 }
]]>
2、A-C-BQ这U情况考虑漏了Q就是通过Cq个U把俩个不连通的集合l打通了
具体代码应该是这?br />for(int j=0;j<now;j++)
if(find(now)!=find(j)&&cross(now,j))
set[find(now)]=find(j);
开始糊涂蛋的写成了
for(int j=0;j<now;j++)
if(cross(now,j))
set[now]=find(j),break;
q个是只考虑了第一U情c?br />
关于数字识别的问题就好办多了Q主要根据部分相交和点相交的数目以及联通集中线D늚数目来判断数字（5?q要Ҏ叉积再来弄下Q，q题受益最q是q查集那里，Ҏ犯的错误?br />
代码如下Q?br />
1 Source Code
2
3 Problem: 3943 User: yzhw
4 Memory: 656K Time: 188MS
5 Language: G++ Result: Accepted
6 Source Code
7 # include <cstdio>
8 # include <vector>
9 # include <cstring>
10 # include <map>
11 using namespace std;
12 # define N 1005
13 # define min(a,b) ((a)<(b)?(a):(b))
14 # define max(a,b) ((a)>(b)?(a):(b))
15 struct line
16 {
17 int x1,y1,x2,y2;
18 }l[N];
19 bool in(int x,int y,const line &pos)
20 {
21 if(x>=min(pos.x1,pos.x2)&&x<=max(pos.x1,pos.x2)&&
22 y>=min(pos.y1,pos.y2)&&y<=max(pos.y1,pos.y2)&&
23 (pos.y2-pos.y1)*(pos.x2-x)==(pos.x2-pos.x1)*(pos.y2-y))
24 return true;
25 else return false;
26 }
27 int cross(const line &a,const line &b)
28 {
29 if(a.x1==b.x1&&a.y1==b.y1||
30 a.x2==b.x1&&a.y2==b.y1||
31 a.x1==b.x2&&a.y1==b.y2||
32 a.x2==b.x2&&a.y2==b.y2) return 1;
33 else if(in(a.x1,a.y1,b)||in(a.x2,a.y2,b)||in(b.x1,b.y1,a)||in(b.x2,b.y2,a)) return 2;
34 else return 0;
35 }
36 int cross(int x1,int y1,int x2,int y2)
37 {
38 return x1*y2-x2*y1;
39 }
40 struct node
41 {
42 vector<line> data;
43 int type() const
44 {
45 int num[3]={0,0,0};
46 for(int i=0;i<data.size();i++)
47 for(int j=i+1;j<data.size();j++)
48 num[cross(data[i],data[j])]++;
49 if(num[1]==4&&num[2]==0)
50 {
51 if(data.size()==4) return 0;
52 else
53 {
54 int target=-1,x1,y1,x2,y2;
55 for(int i=0;i<data.size();i++)
56 {
57 bool flag=true;
58 for(int j=0;j<data.size()&&flag;j++)
59 if(j!=i&&(data[i].x1==data[j].x1&&data[i].y1==data[j].y1||data[i].x1==data[j].x2&&data[i].y1==data[j].y2))
60 flag=false;
61 if(flag)
62 {
63 target=i;
64 x1=data[i].x2,y1=data[i].y2;
65 x2=data[i].x1,y2=data[i].y1;
66 break;
67 }
68 flag=true;
69 for(int j=0;j<data.size()&&flag;j++)
70 if(j!=i&&(data[i].x2==data[j].x1&&data[i].y2==data[j].y1||data[i].x2==data[j].x2&&data[i].y2==data[j].y2))
71 flag=false;
72 if(flag)
73 {
74 target=i;
75 x1=data[i].x1,y1=data[i].y1;
76 x2=data[i].x2,y2=data[i].y2;
77 break;
78 }
79 }
80 for(int i=0;i<data.size();i++)
81 if(i!=target)
82 {
83 if(data[i].x1==x1&&data[i].y1==y1)
84 return cross(data[i].x2-data[i].x1,data[i].y2-data[i].y1,x1-x2,y1-y2)<0?5:2;
85 else if(data[i].x2==x1&&data[i].y2==y1)
86 return cross(data[i].x1-data[i].x2,data[i].y1-data[i].y2,x1-x2,y1-y2)<0?5:2;
87 }
88 }
89 printf("wa!\n");
90 while(true);
91 }
92 else if(num[1]==0&&num[2]==0)
93 return 1;
94 else if(num[1]==4&&num[2]==0)
95 return 2;
96 else if(num[1]==2&&num[2]==1)
97 return 3;
98 else if(num[1]==1&&num[2]==1)
99 return 4;
100 else if(num[1]==4&&num[2]==1)
101 return 6;
102 else if(num[1]==2&&num[2]==0)
103 return 7;
104 else if(num[1]==4&&num[2]==2)
105 return 8;
106 else if(num[1]==3&&num[2]==1)
107 return 9;
108 }
109 };
110 map<int,node> data;
111 int n,set[N],ans[10];
112 int find(int pos)
113 {
114 if(set[pos]==pos) return pos;
115 else return set[pos]=find(set[pos]);
116 }
117 int main()
118 {
119 while(scanf("%d",&n)!=EOF&&n)
120 {
121 data.clear();
122 memset(ans,0,sizeof(ans));
123 for(int i=0;i<n;i++)
124 {
125 set[i]=i;
126 scanf("%d%d%d%d",&l[i].x1,&l[i].y1,&l[i].x2,&l[i].y2);
127 for(int j=0;j<i;j++)
128 if(cross(l[i],l[j])&&find(i)!=find(j))
129 set[find(i)]=find(j);
130 }
131 for(int i=0;i<n;i++)
132 data[find(i)].data.push_back(l[i]);
133 for(map<int,node>::iterator i=data.begin();i!=data.end();i++)
134 ans[i->second.type()]++;
135 printf("%d",ans[0]);
136 for(int i=1;i<10;i++)
137 printf(" %d",ans[i]);
138 printf("\n");
139 }
140 return 0;
141 }
2
3 Problem: 3943 User: yzhw
4 Memory: 656K Time: 188MS
5 Language: G++ Result: Accepted
6 Source Code
7 # include <cstdio>
8 # include <vector>
9 # include <cstring>
10 # include <map>
11 using namespace std;
12 # define N 1005
13 # define min(a,b) ((a)<(b)?(a):(b))
14 # define max(a,b) ((a)>(b)?(a):(b))
15 struct line
16 {
17 int x1,y1,x2,y2;
18 }l[N];
19 bool in(int x,int y,const line &pos)
20 {
21 if(x>=min(pos.x1,pos.x2)&&x<=max(pos.x1,pos.x2)&&
22 y>=min(pos.y1,pos.y2)&&y<=max(pos.y1,pos.y2)&&
23 (pos.y2-pos.y1)*(pos.x2-x)==(pos.x2-pos.x1)*(pos.y2-y))
24 return true;
25 else return false;
26 }
27 int cross(const line &a,const line &b)
28 {
29 if(a.x1==b.x1&&a.y1==b.y1||
30 a.x2==b.x1&&a.y2==b.y1||
31 a.x1==b.x2&&a.y1==b.y2||
32 a.x2==b.x2&&a.y2==b.y2) return 1;
33 else if(in(a.x1,a.y1,b)||in(a.x2,a.y2,b)||in(b.x1,b.y1,a)||in(b.x2,b.y2,a)) return 2;
34 else return 0;
35 }
36 int cross(int x1,int y1,int x2,int y2)
37 {
38 return x1*y2-x2*y1;
39 }
40 struct node
41 {
42 vector<line> data;
43 int type() const
44 {
45 int num[3]={0,0,0};
46 for(int i=0;i<data.size();i++)
47 for(int j=i+1;j<data.size();j++)
48 num[cross(data[i],data[j])]++;
49 if(num[1]==4&&num[2]==0)
50 {
51 if(data.size()==4) return 0;
52 else
53 {
54 int target=-1,x1,y1,x2,y2;
55 for(int i=0;i<data.size();i++)
56 {
57 bool flag=true;
58 for(int j=0;j<data.size()&&flag;j++)
59 if(j!=i&&(data[i].x1==data[j].x1&&data[i].y1==data[j].y1||data[i].x1==data[j].x2&&data[i].y1==data[j].y2))
60 flag=false;
61 if(flag)
62 {
63 target=i;
64 x1=data[i].x2,y1=data[i].y2;
65 x2=data[i].x1,y2=data[i].y1;
66 break;
67 }
68 flag=true;
69 for(int j=0;j<data.size()&&flag;j++)
70 if(j!=i&&(data[i].x2==data[j].x1&&data[i].y2==data[j].y1||data[i].x2==data[j].x2&&data[i].y2==data[j].y2))
71 flag=false;
72 if(flag)
73 {
74 target=i;
75 x1=data[i].x1,y1=data[i].y1;
76 x2=data[i].x2,y2=data[i].y2;
77 break;
78 }
79 }
80 for(int i=0;i<data.size();i++)
81 if(i!=target)
82 {
83 if(data[i].x1==x1&&data[i].y1==y1)
84 return cross(data[i].x2-data[i].x1,data[i].y2-data[i].y1,x1-x2,y1-y2)<0?5:2;
85 else if(data[i].x2==x1&&data[i].y2==y1)
86 return cross(data[i].x1-data[i].x2,data[i].y1-data[i].y2,x1-x2,y1-y2)<0?5:2;
87 }
88 }
89 printf("wa!\n");
90 while(true);
91 }
92 else if(num[1]==0&&num[2]==0)
93 return 1;
94 else if(num[1]==4&&num[2]==0)
95 return 2;
96 else if(num[1]==2&&num[2]==1)
97 return 3;
98 else if(num[1]==1&&num[2]==1)
99 return 4;
100 else if(num[1]==4&&num[2]==1)
101 return 6;
102 else if(num[1]==2&&num[2]==0)
103 return 7;
104 else if(num[1]==4&&num[2]==2)
105 return 8;
106 else if(num[1]==3&&num[2]==1)
107 return 9;
108 }
109 };
110 map<int,node> data;
111 int n,set[N],ans[10];
112 int find(int pos)
113 {
114 if(set[pos]==pos) return pos;
115 else return set[pos]=find(set[pos]);
116 }
117 int main()
118 {
119 while(scanf("%d",&n)!=EOF&&n)
120 {
121 data.clear();
122 memset(ans,0,sizeof(ans));
123 for(int i=0;i<n;i++)
124 {
125 set[i]=i;
126 scanf("%d%d%d%d",&l[i].x1,&l[i].y1,&l[i].x2,&l[i].y2);
127 for(int j=0;j<i;j++)
128 if(cross(l[i],l[j])&&find(i)!=find(j))
129 set[find(i)]=find(j);
130 }
131 for(int i=0;i<n;i++)
132 data[find(i)].data.push_back(l[i]);
133 for(map<int,node>::iterator i=data.begin();i!=data.end();i++)
134 ans[i->second.type()]++;
135 printf("%d",ans[0]);
136 for(int i=1;i<10;i++)
137 printf(" %d",ans[i]);
138 printf("\n");
139 }
140 return 0;
141 }
]]>
]]>
Q?img src="http://acm.hdu.edu.cn/data/images/3682-1.gif" alt="" />
开始写的时候出CBUGQ无奈，上网N解，更无奈，都是些几句话hash然后是一堆难懂的代码。?br />后来仔细想了惻I把重复的操作去除后（是两次删除的同一个木条）Q下面就是个很简单的Ҏ原理?br />因ؓ去除了重复操作，一个木块最多被删除3ơ，然后删除的个数就删除臛_一ơ的个数-删除臛_两次的个?删除臛_3ơ的个数。不能强行枚举，可以用map或者传说中的hash记录被删除掉木块的次数。这里，׃操作最多m=1000Q删除木块数最多ؓC(m,2)Q然后两两枚举操作，把相交木块删除次?1Q然后最后map中所有木块删除次数只能有2个|1?Q当gؓ1Ӟtotal-1,gؓ3Ӟtotal-2
Z么？因ؓ我说了，一个木块最多被删除3ơ，然后俩俩枚D的时候，你懂的?br />
1 # include <cstdio>
2 # include <utility>
3 # include <cstring>
4 # include <algorithm>
5 # include <functional>
6 # include <set>
7 # include <cstdlib>
8 # include <map>
9 using namespace std;
10 struct node
11 {
12 int p[3];
13 bool operator<(const node &pos) const
14 {
15 for(int i=0;i<3;i++)
16 if(p[i]!=pos.p[i])
17 return p[i]<pos.p[i];
18 return false;
19 }
20 };
21 pair<int,int> data[1000][2];
22 set<pair<pair<int,int>,pair<int,int> > > r1;
23 map<node,int> r2;
24 int main()
25 {
26 int test;
27 scanf("%d",&test);
28 while(test--)
29 {
30 int n,m,p=0;
31 char str[128];
32 scanf("%d%d",&n,&m);
33 r1.clear();r2.clear();
34 while(m--)
35 {
36 scanf("%s",str);
37 char *t=strtok(str,",");
38 data[p][0].first=(*t)-'X';
39 data[p][0].second=atoi(t+2);
40 t=strtok(NULL," ");
41 data[p][1].first=(*t)-'X';
42 data[p][1].second=atoi(t+2);
43 if(data[p][0].first>data[p][1].first) swap(data[p][0],data[p][1]);
44 pair< pair<int,int>,pair<int,int> > tt;
45 tt.first=data[p][0];
46 tt.second=data[p][1];
47 if(data[p][0].second>=1&&data[p][0].second<=n&&data[p][1].second>=1&&data[p][1].second<=n&&r1.find(tt)==r1.end()) p++,r1.insert(tt);
48 }
49 m=p;
50 for(int i=0;i<m;i++)
51 for(int j=i+1;j<m;j++)
52 if(data[i][0]==data[j][0]&&data[i][1].first!=data[j][1].first||
53 data[i][1]==data[j][0]&&data[i][0].first!=data[j][1].first||
54 data[i][0]==data[j][1]&&data[i][1].first!=data[j][0].first||
55 data[i][1]==data[j][1]&&data[i][0].first!=data[j][0].first)
56 {
57 node t;
58 t.p[data[i][0].first]=data[i][0].second;
59 t.p[data[i][1].first]=data[i][1].second;
60 t.p[data[j][0].first]=data[j][0].second;
61 t.p[data[j][1].first]=data[j][1].second;
62 r2[t]++;
63 }
64 int total=m*n;
65 for(map<node,int>::iterator i=r2.begin();i!=r2.end();i++)
66 if(i->second==1) total--;
67 else total-=2;
68 printf("%d\n",total);
69 }
70 //system("pause");
71 return 0;
72 }
73
2 # include <utility>
3 # include <cstring>
4 # include <algorithm>
5 # include <functional>
6 # include <set>
7 # include <cstdlib>
8 # include <map>
9 using namespace std;
10 struct node
11 {
12 int p[3];
13 bool operator<(const node &pos) const
14 {
15 for(int i=0;i<3;i++)
16 if(p[i]!=pos.p[i])
17 return p[i]<pos.p[i];
18 return false;
19 }
20 };
21 pair<int,int> data[1000][2];
22 set<pair<pair<int,int>,pair<int,int> > > r1;
23 map<node,int> r2;
24 int main()
25 {
26 int test;
27 scanf("%d",&test);
28 while(test--)
29 {
30 int n,m,p=0;
31 char str[128];
32 scanf("%d%d",&n,&m);
33 r1.clear();r2.clear();
34 while(m--)
35 {
36 scanf("%s",str);
37 char *t=strtok(str,",");
38 data[p][0].first=(*t)-'X';
39 data[p][0].second=atoi(t+2);
40 t=strtok(NULL," ");
41 data[p][1].first=(*t)-'X';
42 data[p][1].second=atoi(t+2);
43 if(data[p][0].first>data[p][1].first) swap(data[p][0],data[p][1]);
44 pair< pair<int,int>,pair<int,int> > tt;
45 tt.first=data[p][0];
46 tt.second=data[p][1];
47 if(data[p][0].second>=1&&data[p][0].second<=n&&data[p][1].second>=1&&data[p][1].second<=n&&r1.find(tt)==r1.end()) p++,r1.insert(tt);
48 }
49 m=p;
50 for(int i=0;i<m;i++)
51 for(int j=i+1;j<m;j++)
52 if(data[i][0]==data[j][0]&&data[i][1].first!=data[j][1].first||
53 data[i][1]==data[j][0]&&data[i][0].first!=data[j][1].first||
54 data[i][0]==data[j][1]&&data[i][1].first!=data[j][0].first||
55 data[i][1]==data[j][1]&&data[i][0].first!=data[j][0].first)
56 {
57 node t;
58 t.p[data[i][0].first]=data[i][0].second;
59 t.p[data[i][1].first]=data[i][1].second;
60 t.p[data[j][0].first]=data[j][0].second;
61 t.p[data[j][1].first]=data[j][1].second;
62 r2[t]++;
63 }
64 int total=m*n;
65 for(map<node,int>::iterator i=r2.begin();i!=r2.end();i++)
66 if(i->second==1) total--;
67 else total-=2;
68 printf("%d\n",total);
69 }
70 //system("pause");
71 return 0;
72 }
73
]]>
]]>
1、插?br />按照划分树的定义Q如果小于有序表中中间节点的|递归插入左子树，否则递归插入叛_树。更新当前区间段的划分信息（无非是往后计一个）
2、询问s,e区间Wk数
查询s,e区间里面划分到左子树的个数iQ如果i>=kQ那么显焉归到左子树查询Q否则就是递归到右子树查询k-i的数。注意！q里要重新定位左子树和右子树中的区间Q由于是闭区_那么做端点ؓs+sum(s-1),右端点ؓs+sum(e)-1Q这个减一丢了。。调了我半天。。哎。。以前写都是左闭叛_区间的，l果习惯了。?br />3、查询gؓk的数的位?br />q个需要一个辅助数l，记录gؓk的数插在最层区间的哪个位|了。这个办好后Q就Ҏ了，如果数被划分C左子树，那么递归查询左子树，否则q回递归查询叛_树的值加上当前区间被划分到左子树的个?br />4、查询rank k的数
同样是这P如果当前区间被划分到左子树的个数于{于kQ那么递归查询左子树，否则递归查询叛_树中rank为k-左子树的size?br />大概思想是q样了，实现有很多细节，比如说假设p==区间左端点（左区间木有数Q，那么sum(p-1)的时候就要特判下了。我喜欢用三元式Q很方便?br />
代码
# include <cstdio>
# include <cstring>
# include <map>
using namespace std;
# define N 100005
int arr[20][N];
struct node
{
int s,e,layer;
int c;
}st[4*N];
int q[500000][4],c;
int remap[N],position[N];
map<int,int> refer;
void init(int s,int e,int layer,int pos=1)
{
st[pos].s=s;st[pos].e=e;
st[pos].layer=layer;
st[pos].c=st[pos].s;
if(s!=e) init(s,(s+e)/2,layer+1,pos<<1),init((s+e)/2+1,e,layer+1,(pos<<1)+1);
}
void insert(int value,int pos=1)
{
if(st[pos].s==st[pos].e)
arr[st[pos].layer][st[pos].c++]=value;
else
{
if(value<=(st[pos].s+st[pos].e)/2)
{
arr[st[pos].layer][st[pos].c]=(st[pos].c==st[pos].s?0:arr[st[pos].layer][st[pos].c-1])+1;
st[pos].c++;
insert(value,pos<<1);
}
else
{
arr[st[pos].layer][st[pos].c]=(st[pos].c==st[pos].s?0:arr[st[pos].layer][st[pos].c-1]);
st[pos].c++;
insert(value,(pos<<1)+1);
}
}
}
int q1(int s,int t,int k,int pos=1)
{
if(st[pos].s==st[pos].e)
return remap[arr[st[pos].layer][st[pos].s]];
else
{
if(arr[st[pos].layer][t]-(s==st[pos].s?0:arr[st[pos].layer][s-1])>=k)//left
return q1(st[pos].s+(s==st[pos].s?0:arr[st[pos].layer][s-1]),st[pos].s+arr[st[pos].layer][t]-1,k,pos<<1);
else//right
{
k-=arr[st[pos].layer][t]-(s==st[pos].s?0:arr[st[pos].layer][s-1]);
return q1((st[pos].s+st[pos].e)/2+1+s-1-st[pos].s+1-(s==st[pos].s?0:arr[st[pos].layer][s-1]),(st[pos].s+st[pos].e)/2+1+t-st[pos].s+1-arr[st[pos].layer][t]-1,k,(pos<<1)+1);
}
}
}
int q2(int s,int pos=1)
{
if(st[pos].s==st[pos].e) return 1;
else if(arr[st[pos].layer][s]-(s==st[pos].s?0:arr[st[pos].layer][s-1]))
return q2(st[pos].s+arr[st[pos].layer][s]-1,pos<<1);
else
return (st[pos].c==st[pos].s?0:arr[st[pos].layer][st[pos].c-1])+q2((st[pos].s+st[pos].e)/2+1+s-st[pos].s+1-arr[st[pos].layer][s]-1,(pos<<1)+1);
}
int q3(int k,int pos=1)
{
if(st[pos].s==st[pos].e) return remap[arr[st[pos].layer][st[pos].s]];
else if(k<=(st[pos].c==st[pos].s?0:arr[st[pos].layer][st[pos].c-1])) return q3(k,pos<<1);
else return q3(k-(st[pos].s==st[pos].c?0:arr[st[pos].layer][st[pos].c-1]),(pos<<1)+1);
}
int main()
{
int n,test=1;
while(scanf("%d",&n)!=EOF)
{
refer.clear();c=1;
memset(arr,0,sizeof(arr));
for(int i=0;i<n;i++)
{
char tmp[12];
scanf("%s",tmp);
if(*tmp=='I')
q[i][0]=0;
else q[i][0]=tmp[6]-48;
switch(q[i][0])
{
case 0:
scanf("%d",&q[i][1]);
refer[q[i][1]]=0;
break;
case 1:
scanf("%d%d%d",&q[i][1],&q[i][2],&q[i][3]);
break;
default:
scanf("%d",&q[i][1]);
break;
};
}
for(map<int,int>::iterator i=refer.begin();i!=refer.end();i++)
remap[c]=i->first,i->second=c++;
init(1,c-1,0);
long long t[4]={0,0,0,0};
int now=1;
for(int i=0;i<n;i++)
switch(q[i][0])
{
case 0:
insert(refer[q[i][1]]);
position[refer[q[i][1]]]=now++;
break;
case 1:
t[1]+=q1(q[i][1],q[i][2],q[i][3]);
break;
case 2:
t[2]+=q2(position[refer[q[i][1]]]);
break;
case 3:
t[3]+=q3(q[i][1]);
break;
};
printf("Case %d:\n%I64d\n%I64d\n%I64d\n",test++,t[1],t[2],t[3]);
}
return 0;
}
]]>