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 }
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璁＄畻綾諱技榪欐牱涓涓浘褰㈢殑闈㈢Н銆佽竟涓婄殑鏍肩偣鏁般佸唴閮ㄦ牸鐐規(guī)暟
瑙ｆ硶錛?br>榪欓噷鐢ㄥ埌涓涓畾鐞嗭紝鍙玴ick瀹氱悊
闈㈢Н=杈逛笂鐐規(guī)暟/2-1+鍐呴儴鐐規(guī)暟
鐒跺悗姹傝竟涓婄殑鐐規(guī)暟鐩存帴鐢╣cd(dx,dy)灝卞彲浠ヤ簡(jiǎn)銆?br>緗戞牸鍥炬槸涓涓濂囩殑鍥撅紝閲岄潰鏈夊緢澶氳寮傜殑緇撹銆?br>
浠ｇ爜錛?br>
1
# include <stdio.h>
2# define cross(x1,y1,x2,y2) ((x1)*(y2)-(x2)*(y1))
3int p[105][2];
4int gcd(int n1,int n2)
5
{
6
if(n1<0) n1*=-1;
7 if(n2<0) n2*=-1;
8 while(n2)
9
{
10 int t=n1%n2;
11 n1=n2;
12 n2=t;
13
}
14 return n1;
15
}
16int main()
17{
18 //freopen("ans.txt","w",stdout);
19 int test,t;
20 scanf("%d",&test);
21 for(t=1;t<=test;t++)
22 {
23 int n,i;
24 int aera=0,edge=0;
25 scanf("%d",&n);
26 for(i=1;i<=n;i++)
27 {
28 scanf("%d%d",&p[i][0],&p[i][1]);
29 edge+=gcd(p[i][0],p[i][1]);
30 }
31 p[0][0]=p[0][1]=0;
32 for(i=1;i<n;i++)
33 p[i][0]+=p[i-1][0],p[i][1]+=p[i-1][1];
34 for(i=2;i<n;i++)
35 aera+=cross(p[i-1][0],p[i-1][1],p[i][0],p[i][1]);
36 printf("Scenario #%d:\n%d %d %.1f\n\n",t,(int)((aera+2-edge)*0.5+1e-6),edge,aera*0.5);
37 }
38 // system("pause");
39 return 0;
40}
41
# include <stdio.h>2
# define cross(x1,y1,x2,y2) ((x1)*(y2)-(x2)*(y1))3
int p[105][2];4
int gcd(int n1,int n2)5
{6
if(n1<0) n1*=-1;7
if(n2<0) n2*=-1;8
while(n2)9
{10
int t=n1%n2;11
n1=n2;12
n2=t;13
}14
return n1;15
}16
int main()17
{18
//freopen("ans.txt","w",stdout);19
int test,t;20
scanf("%d",&test);21
for(t=1;t<=test;t++)22
{23
int n,i;24
int aera=0,edge=0;25
scanf("%d",&n);26
for(i=1;i<=n;i++)27
{28
scanf("%d%d",&p[i][0],&p[i][1]);29
edge+=gcd(p[i][0],p[i][1]);30
}31
p[0][0]=p[0][1]=0;32
for(i=1;i<n;i++)33
p[i][0]+=p[i-1][0],p[i][1]+=p[i-1][1];34
for(i=2;i<n;i++)35
aera+=cross(p[i-1][0],p[i-1][1],p[i][0],p[i][1]);36
printf("Scenario #%d:\n%d %d %.1f\n\n",t,(int)((aera+2-edge)*0.5+1e-6),edge,aera*0.5); 37
}38
// system("pause");39
return 0;40
}41
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1# include <iostream>
2# include <cmath>
3using namespace std;
4# include <set>
5struct cmp
6{
7 bool operator()(const double a,const double b) const
8 {
9 return fabs(a-b)>1e-8&&a<b;
10 }
11};
12set<double,cmp> refer;
13bool sp=0;
14int data[201][2];
15int main()
16{
17 int n;
18 cin>>n;
19 for(int i=0;i<n;i++)
20 {
21 cin>>data[i][0]>>data[i][1];
22 for(int j=0;j<i;j++)
23 if(data[j][0]==data[i][0])
24 sp=1;
25 else
26 refer.insert((data[j][1]-data[i][1])/(double)(data[j][0]-data[i][0]));
27 }
28 cout<<refer.size()+sp<<endl;
29 // system("pause");
30 return 0;
31}
32
# include <iostream>2
# include <cmath>3
using namespace std;4
# include <set>5
struct cmp6
{7
bool operator()(const double a,const double b) const8
{9
return fabs(a-b)>1e-8&&a<b;10
}11
}; 12
set<double,cmp> refer;13
bool sp=0;14
int data[201][2];15
int main()16
{17
int n;18
cin>>n;19
for(int i=0;i<n;i++)20
{21
cin>>data[i][0]>>data[i][1];22
for(int j=0;j<i;j++)23
if(data[j][0]==data[i][0])24
sp=1;25
else26
refer.insert((data[j][1]-data[i][1])/(double)(data[j][0]-data[i][0]));27
}28
cout<<refer.size()+sp<<endl;29
// system("pause");30
return 0;31
}32
]]>
緇欏嚭涓涓湁n(n<=100)涓偣鐨勭畝鍗曞杈瑰艦錛屾眰璇ュ杈瑰艦鐨勯噸蹇?jī)銆?
Solution
鏍規(guī)嵁澶氳竟褰㈤噸蹇?jī)鐨勫畾涔夊Q屽彲浠ュ鍏惰繘琛屼笁瑙掑墫鍒嗭紝璁＄畻姣忎釜涓夎褰㈢殑闈㈢Н浠ュ強(qiáng)閲嶅績(jī)銆傚杈瑰艦鐨勯噸蹇?jī)灏辨槸鎵鏈変笁瑙掑艦鐨勯噸蹇?jī)瀵归潰绉殑鍔犳潈邈^鍧囨暟錛屼篃灝辨槸璇達(dá)細(xì)
center.x = (cen[0].x * area[0] + cen[1].x * area[1] ..... + cen[n].x * area[n]) / totalarea
center.y = (cen[0].y * area[0] + cen[1].y * area[1] ..... + cen[n].y * area[n]) / totalarea
cen[i]浠ｈ〃絎琲涓笁瑙掑艦鐨勯噸蹇?jī)锛屼笁瑙掑舰鐨勯噸蹇?jī)?yōu)鏄Q?
center_of_tri.x=(p1.x+p2.x+p3.x)/3.0
center_of_tri.y=(p1.y+p2.y+p3.y)/3.0
area[i]灝辨槸絎琲涓笁瑙掑艦鐨勯潰縐倀otalarea灝辨槸澶氳竟褰㈢殑鎬婚潰縐?
1#include <cstdio>
2#include <cmath>
3#define EPS 1e-8
4#define N 105
5struct POINT {
6 double x, y;
7 POINT() {
8 x = y = 0;
9 }
10 POINT(double x, double y) :
11 x(x), y(y) {
12 }
13 void get() {
14 scanf("%lf%lf", &x, &y);
15 }
16 void print() {
17 printf("%.6lf %.6lf\n", x, y);
18 }
19 POINT operator+(const POINT &p) {
20 return POINT(x + p.x, y + p.y);
21 }
22};
23POINT pl[N];
24int n;
25double cross(POINT o, POINT &a, POINT &b) {
26 return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
27}
28int sgn(double x) {
29 return x < -EPS ? -1 : x > EPS;
30}
31void solve() {
32 int i;
33 double area = 0;
34 POINT ct;
35 for (i = 0; i < n; i++)
36 pl[i].get();
37 pl[i] = pl[0];
38 for (i = 0; i < n; i++) {
39 double s = cross(POINT(), pl[i], pl[i + 1]);
40 POINT t;
41 area += s;
42 t = pl[i] + pl[i + 1];
43 ct.x += s * t.x, ct.y += s * t.y;
44 }
45 ct.x = ct.x / area / 3.0, ct.y = ct.y / area / 3.0;
46 ct.print();
47}
48int main() {
49 int T = 1;
50 while (scanf("%d", &n) && n) {
51 printf("Stage #%d: ", T++);
52 solve();
53 }
54 return 0;
55}
56
#include <cstdio>2
#include <cmath>3
#define EPS 1e-84
#define N 1055
struct POINT {6
double x, y;7
POINT() {8
x = y = 0;9
}10
POINT(double x, double y) :11
x(x), y(y) {12
}13
void get() {14
scanf("%lf%lf", &x, &y);15
}16
void print() {17
printf("%.6lf %.6lf\n", x, y);18
}19
POINT operator+(const POINT &p) {20
return POINT(x + p.x, y + p.y);21
}22
};23
POINT pl[N];24
int n;25
double cross(POINT o, POINT &a, POINT &b) {26
return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);27
}28
int sgn(double x) {29
return x < -EPS ? -1 : x > EPS;30
}31
void solve() {32
int i;33
double area = 0;34
POINT ct;35
for (i = 0; i < n; i++)36
pl[i].get();37
pl[i] = pl[0];38
for (i = 0; i < n; i++) {39
double s = cross(POINT(), pl[i], pl[i + 1]);40
POINT t;41
area += s;42
t = pl[i] + pl[i + 1];43
ct.x += s * t.x, ct.y += s * t.y;44
}45
ct.x = ct.x / area / 3.0, ct.y = ct.y / area / 3.0;46
ct.print();47
}48
int main() {49
int T = 1;50
while (scanf("%d", &n) && n) {51
printf("Stage #%d: ", T++);52
solve();53
}54
return 0;55
}56
]]>