MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋 
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1098
題目分析:
純粹的數(shù)學(xué)題, 數(shù)學(xué)歸納法的 應(yīng)用 , 最后 歸納得出 原題等價與 18 + k*a 是否能被65整除.
代碼如下 :
代碼/*
Mail to : miyubai@gamil.com
My Blog : www.baiyun.me
Link : http://www.cnblogs.com/MiYu || http://m.shnenglu.com/MiYu
Author By : MiYu
Test : 1
Complier : g++ mingw32-3.4.2
Program : HDU_1098
Doc Name : Ignatius's puzzle
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
inline bool scan_d(int &num)
{
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<'0'||in>'9')) in=getchar();
if(in=='-'){ IsN=true;num=0;}
else num=in-'0';
while(in=getchar(),in>='0'&&in<='9'){
num*=10,num+=in-'0';
}
if(IsN) num=-num;
return true;
}
int main ()
{
int N;
while ( scan_d ( N ) ) {
if ( N % 65 == 0 )
puts ( "no" );
else {
int i;
for ( i = 0; i < 65; ++ i ) {
if ( ( i * N + 18 ) % 65 == 0 ) {
printf ( "%d\n", i );
break;
}
}
if ( i == 65 ) puts ( "no" );
}
}
return 0;
}
Mail to : miyubai@gamil.com
My Blog : www.baiyun.me
Link : http://www.cnblogs.com/MiYu || http://m.shnenglu.com/MiYu
Author By : MiYu
Test : 1
Complier : g++ mingw32-3.4.2
Program : HDU_1098
Doc Name : Ignatius's puzzle
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
inline bool scan_d(int &num)
{
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<'0'||in>'9')) in=getchar();
if(in=='-'){ IsN=true;num=0;}
else num=in-'0';
while(in=getchar(),in>='0'&&in<='9'){
num*=10,num+=in-'0';
}
if(IsN) num=-num;
return true;
}
int main ()
{
int N;
while ( scan_d ( N ) ) {
if ( N % 65 == 0 )
puts ( "no" );
else {
int i;
for ( i = 0; i < 65; ++ i ) {
if ( ( i * N + 18 ) % 65 == 0 ) {
printf ( "%d\n", i );
break;
}
}
if ( i == 65 ) puts ( "no" );
}
}
return 0;
}
]]>
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1999
題目描述:
不可摸數(shù)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2613 Accepted Submission(s): 694
Problem Description
s(n)是正整數(shù)n的真因子之和，即小于n且整除n的因子和.例如s(12)=1+2+3+4+6=16.如果任何
數(shù)m，s(m)都不等于n,則稱n為不可摸數(shù).
Input
包含多組數(shù)據(jù)，首先輸入T,表示有T組數(shù)據(jù).每組數(shù)據(jù)1行給出n(2<=n<=1000)是整數(shù)。
Output
如果n是不可摸數(shù)，輸出yes，否則輸出no
Sample Input
3
2
5
8
Sample Output
yes
yes
no
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2613 Accepted Submission(s): 694
Problem Description
s(n)是正整數(shù)n的真因子之和，即小于n且整除n的因子和.例如s(12)=1+2+3+4+6=16.如果任何
數(shù)m，s(m)都不等于n,則稱n為不可摸數(shù).
Input
包含多組數(shù)據(jù)，首先輸入T,表示有T組數(shù)據(jù).每組數(shù)據(jù)1行給出n(2<=n<=1000)是整數(shù)。
Output
如果n是不可摸數(shù)，輸出yes，否則輸出no
Sample Input
3
2
5
8
Sample Output
yes
yes
no
題目分析:
標(biāo)準(zhǔn)的篩選法。求出每個數(shù)的因子和，
然后看因子和是否在1000以內(nèi)，是的話就證明等于因子和的這個數(shù)是不可摸數(shù)。
然后看因子和是否在1000以內(nèi)，是的話就證明等于因子和的這個數(shù)是不可摸數(shù)。
代碼如下: ( 奮斗哥代碼 0rz................... )
#include <iostream>
#include <string.h>
#include <cmath>
using namespace std;
int sum[1000001], sign[1001];
int main()
{
int nCases, num;
scanf("%d", &nCases);
for(int i = 1; i <= 500000; ++i)
for(int j = 2*i; j <= 1000000; j += i)
sum[j] += i;
for(int i = 1; i <= 1000000; ++i)
if(sum[i] < 1000)
sign[sum[i]] = 1;
while(nCases--)
{
scanf("%d", &num);
if(sign[num])
printf("no\n");
else
printf("yes\n");
}
return 0;
}
#include <string.h>
#include <cmath>
using namespace std;
int sum[1000001], sign[1001];
int main()
{
int nCases, num;
scanf("%d", &nCases);
for(int i = 1; i <= 500000; ++i)
for(int j = 2*i; j <= 1000000; j += i)
sum[j] += i;
for(int i = 1; i <= 1000000; ++i)
if(sum[i] < 1000)
sign[sum[i]] = 1;
while(nCases--)
{
scanf("%d", &num);
if(sign[num])
printf("no\n");
else
printf("yes\n");
}
return 0;
}
]]>
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2036
題目描述:
改革春風(fēng)吹滿地
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5623 Accepted Submission(s): 2763
Problem Description
“ 改革春風(fēng)吹滿地,
不會AC沒關(guān)系;
實在不行回老家，
還有一畝三分地。
謝謝!（樂隊奏樂）”
話說部分學(xué)生心態(tài)極好，每天就知道游戲，這次考試如此簡單的題目，也是云里霧里，而且，還竟然來這么幾句打油詩。
好呀，老師的責(zé)任就是幫你解決問題，既然想種田，那就分你一塊。
這塊田位于浙江省溫州市蒼南縣靈溪鎮(zhèn)林家鋪子村，多邊形形狀的一塊地，原本是linle 的，現(xiàn)在就準(zhǔn)備送給你了。不過，任何事情都沒有那么簡單，你必須首先告訴我這塊地到底有多少面積，如果回答正確才能真正得到這塊地。
發(fā)愁了吧？就是要讓你知道，種地也是需要AC知識的！以后還是好好練吧
Input
輸入數(shù)據(jù)包含多個測試實例，每個測試實例占一行，每行的開始是一個整數(shù)n(3<=n<=100)，它表示多邊形的邊數(shù)（當(dāng)然也是頂點數(shù)），然后是按照逆時針順序給出的n個頂點的坐標(biāo)（x1, y1, x2, y2 xn, yn）,為了簡化問題，這里的所有坐標(biāo)都用整數(shù)表示。
輸入數(shù)據(jù)中所有的整數(shù)都在32位整數(shù)范圍內(nèi)，n=0表示數(shù)據(jù)的結(jié)束，不做處理。
Output
對于每個測試實例，請輸出對應(yīng)的多邊形面積，結(jié)果精確到小數(shù)點后一位小數(shù)。
每個實例的輸出占一行。
Sample Input
3 0 0 1 0 0 1
4 1 0 0 1 -1 0 0 -1
0
Sample Output
0.5
2.0
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5623 Accepted Submission(s): 2763
Problem Description
“ 改革春風(fēng)吹滿地,
不會AC沒關(guān)系;
實在不行回老家，
還有一畝三分地。
謝謝!（樂隊奏樂）”
話說部分學(xué)生心態(tài)極好，每天就知道游戲，這次考試如此簡單的題目，也是云里霧里，而且，還竟然來這么幾句打油詩。
好呀，老師的責(zé)任就是幫你解決問題，既然想種田，那就分你一塊。
這塊田位于浙江省溫州市蒼南縣靈溪鎮(zhèn)林家鋪子村，多邊形形狀的一塊地，原本是linle 的，現(xiàn)在就準(zhǔn)備送給你了。不過，任何事情都沒有那么簡單，你必須首先告訴我這塊地到底有多少面積，如果回答正確才能真正得到這塊地。
發(fā)愁了吧？就是要讓你知道，種地也是需要AC知識的！以后還是好好練吧
Input
輸入數(shù)據(jù)包含多個測試實例，每個測試實例占一行，每行的開始是一個整數(shù)n(3<=n<=100)，它表示多邊形的邊數(shù)（當(dāng)然也是頂點數(shù)），然后是按照逆時針順序給出的n個頂點的坐標(biāo)（x1, y1, x2, y2
xn, yn）,為了簡化問題，這里的所有坐標(biāo)都用整數(shù)表示。輸入數(shù)據(jù)中所有的整數(shù)都在32位整數(shù)范圍內(nèi)，n=0表示數(shù)據(jù)的結(jié)束，不做處理。
Output
對于每個測試實例，請輸出對應(yīng)的多邊形面積，結(jié)果精確到小數(shù)點后一位小數(shù)。
每個實例的輸出占一行。
Sample Input
3 0 0 1 0 0 1
4 1 0 0 1 -1 0 0 -1
0
Sample Output
0.5
2.0
題目分析:
模板題. 直接用 多邊形面積公式 : S = 1/2 * abs( ∑(xiyi+1 – xi+1yi) )
代碼如下:
#include <iostream>
#include <cmath>
using namespace std;
typedef struct point{
int x, y;
}point;
point polygon[101];
double areaGmty ( int num )
{
double area = 0;
for ( int i = 0; i < num; ++ i )
{
area += ( polygon[i].x * polygon[(i+1)%num].y ) - (polygon[(i+1)%num].x * polygon[i].y);
}
return area;
}
int main ()
{
int N;
while ( scanf ( "%d",&N ), N )
{
for ( int i = 0; i != N; ++ i )
{
scanf ( "%d%d", &polygon[i].x,&polygon[i].y );
}
double area = areaGmty ( N ) / 2.0;
printf ( "%.1lf\n",area );
}
return 0;
}
#include <cmath>
using namespace std;
typedef struct point{
int x, y;
}point;
point polygon[101];
double areaGmty ( int num )
{
double area = 0;
for ( int i = 0; i < num; ++ i )
{
area += ( polygon[i].x * polygon[(i+1)%num].y ) - (polygon[(i+1)%num].x * polygon[i].y);
}
return area;
}
int main ()
{
int N;
while ( scanf ( "%d",&N ), N )
{
for ( int i = 0; i != N; ++ i )
{
scanf ( "%d%d", &polygon[i].x,&polygon[i].y );
}
double area = areaGmty ( N ) / 2.0;
printf ( "%.1lf\n",area );
}
return 0;
}
]]>
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1969
題目描述:
Pie
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 65
Problem Description
My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.
My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.
What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.
Input
One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.
Output
For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).
Sample Input
3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2
Sample Output
25.1327
3.1416
50.2655
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 65
Problem Description
My birthday is coming up and traditionally I'm serving pie. Not just one pie, no, I have a number N of them, of various tastes and of various sizes. F of my friends are coming to my party and each of them gets a piece of pie. This should be one piece of one pie, not several small pieces since that looks messy. This piece can be one whole pie though.
My friends are very annoying and if one of them gets a bigger piece than the others, they start complaining. Therefore all of them should get equally sized (but not necessarily equally shaped) pieces, even if this leads to some pie getting spoiled (which is better than spoiling the party). Of course, I want a piece of pie for myself too, and that piece should also be of the same size.
What is the largest possible piece size all of us can get? All the pies are cylindrical in shape and they all have the same height 1, but the radii of the pies can be different.
Input
One line with a positive integer: the number of test cases. Then for each test case:
---One line with two integers N and F with 1 <= N, F <= 10 000: the number of pies and the number of friends.
---One line with N integers ri with 1 <= ri <= 10 000: the radii of the pies.
Output
For each test case, output one line with the largest possible volume V such that me and my friends can all get a pie piece of size V. The answer should be given as a floating point number with an absolute error of at most 10^(-3).
Sample Input
3
3 3
4 3 3
1 24
5
10 5
1 4 2 3 4 5 6 5 4 2
Sample Output
25.1327
3.1416
50.2655
題目分析:
2分求解.
題目大意是要辦生日Party，有n個餡餅，有f個朋友，接下來是n個餡餅的半徑。然后是分餡餅了，
注意咯自己也要，大家都要一樣大，形狀沒什么要求，但都要是一整塊的那種，也就是說不能從兩個餅中
各割一小塊來湊一塊，像面積為10的和6的兩塊餅（餅的厚度是1，所以面積和體積相等），
如果每人分到面積為5，則10分兩塊，6切成5，夠分3個人，如果每人6，則只能分兩個了！
題目要求我們分到的餅盡可能的大！
只要注意精度問題就可以了,一般WA 都是精度問題.
題目大意是要辦生日Party，有n個餡餅，有f個朋友，接下來是n個餡餅的半徑。然后是分餡餅了，
注意咯自己也要，大家都要一樣大，形狀沒什么要求，但都要是一整塊的那種，也就是說不能從兩個餅中
各割一小塊來湊一塊，像面積為10的和6的兩塊餅（餅的厚度是1，所以面積和體積相等），
如果每人分到面積為5，則10分兩塊，6切成5，夠分3個人，如果每人6，則只能分兩個了！
題目要求我們分到的餅盡可能的大！
只要注意精度問題就可以了,一般WA 都是精度問題.
代碼如下:
#include <iostream>
#include <cmath>
using namespace std;
double a[10005];
int N,f;
double pi = acos ( -1.0 );
bool test ( double x )
{
int count = 0;
for ( int i = 1; i <= N; ++ i )
{
count += int ( a[i] / x );
}
if ( count >= f + 1 )
{
return true;
}
else
{
return false;
}
}
int main()
{
int T;
scanf ( "%d", &T );
while ( T -- )
{
double sum = 0;
double rad;
scanf("%d%d",&N,&f);
for ( int i = 1; i <= N; ++ i )
{
scanf ( "%lf", &rad );
a[i] = rad * rad * pi;
sum += a[i];
}
double max = sum / ( f + 1 );
double l = 0.0;
double r = max;
double mid = 0.0;
while ( r - l > 1e-6 )
{
mid = ( l + r ) / 2;
if ( test ( mid ) )
{
l = mid;
}
else
{
r = mid;
}
}
printf("%.4lf\n",mid);
}
return 0;
}
#include <cmath>
using namespace std;
double a[10005];
int N,f;
double pi = acos ( -1.0 );
bool test ( double x )
{
int count = 0;
for ( int i = 1; i <= N; ++ i )
{
count += int ( a[i] / x );
}
if ( count >= f + 1 )
{
return true;
}
else
{
return false;
}
}
int main()
{
int T;
scanf ( "%d", &T );
while ( T -- )
{
double sum = 0;
double rad;
scanf("%d%d",&N,&f);
for ( int i = 1; i <= N; ++ i )
{
scanf ( "%lf", &rad );
a[i] = rad * rad * pi;
sum += a[i];
}
double max = sum / ( f + 1 );
double l = 0.0;
double r = max;
double mid = 0.0;
while ( r - l > 1e-6 )
{
mid = ( l + r ) / 2;
if ( test ( mid ) )
{
l = mid;
}
else
{
r = mid;
}
}
printf("%.4lf\n",mid);
}
return 0;
}
]]>
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2899
題目描述:
Strange fuction
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 95 Accepted Submission(s): 75
Problem Description
Now, here is a fuction:
F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)
Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
Sample Input
2
100
200
Sample Output
-74.4291
-178.8534
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 95 Accepted Submission(s): 75
Problem Description
Now, here is a fuction:
F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100)
Can you find the minimum value when x is between 0 and 100.
Input
The first line of the input contains an integer T(1<=T<=100) which means the number of test cases. Then T lines follow, each line has only one real numbers Y.(0 < Y <1e10)
Output
Just the minimum value (accurate up to 4 decimal places),when x is between 0 and 100.
Sample Input
2
100
200
Sample Output
-74.4291
-178.8534
題目分析:
純數(shù)學(xué)題, 需要微積分的知識分析題目...........我承認(rèn)....我數(shù)學(xué)沒怎么學(xué)好..........
分析出來使用2分搜索就可以了..........
分析出來使用2分搜索就可以了..........
代碼如下:
#include <iostream>
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋
#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
double cal ( double n )
{
return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
int main ()
{
int T;
scanf ( "%d",&T );
while ( T -- )
{
scanf ( "%lf",&y );
if ( cal(0) > y || cal(100) < y )
{
printf ( "No solution!\n" );
continue;
}
double l = 0.0, r = 100.0,res = 0.0;
while ( r - l > 1e-6 )
{
double mid = ( l + r ) / 2.0;
res = cal ( mid );
if ( res > y )
r = mid - 1e-6;
else
l = mid + 1e-6;
}
printf ( "%.4lf\n",( l + r ) / 2.0 );
}
return 0;
}
#include <cmath>
using namespace std;
#define POW(x) ( (x) * (x) )
#define POW3(x) ( POW(x) * (x) )
MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋
#define POW4(x) ( POW(x) * POW(x) )
double y = 0;
double cal ( double n )
{
return 8.0 * POW4(n) + 7 * POW3(n) + 2 * POW(n) + 3 * n + 6 ;
}
int main ()
{
int T;
scanf ( "%d",&T );
while ( T -- )
{
scanf ( "%lf",&y );
if ( cal(0) > y || cal(100) < y )
{
printf ( "No solution!\n" );
continue;
}
double l = 0.0, r = 100.0,res = 0.0;
while ( r - l > 1e-6 )
{
double mid = ( l + r ) / 2.0;
res = cal ( mid );
if ( res > y )
r = mid - 1e-6;
else
l = mid + 1e-6;
}
printf ( "%.4lf\n",( l + r ) / 2.0 );
}
return 0;
}
]]>
題目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=2073
題目描述:
Problem Description
甜甜從小就喜歡畫圖畫，最近他買了一支智能畫筆，由于剛剛接觸，所以甜甜只會用它來畫直線，于是他就在平面直角坐標(biāo)系中畫出如下的圖形：
甜甜的好朋友蜜蜜發(fā)現(xiàn)上面的圖還是有點規(guī)則的，于是他問甜甜：在你畫的圖中，我給你兩個點，請你算一算連接兩點的折線長度（即沿折線走的路線長度）吧。
Input
第一個數(shù)是正整數(shù)N（≤100）。代表數(shù)據(jù)的組數(shù)。
每組數(shù)據(jù)由四個非負(fù)整數(shù)組成x1，y1，x2，y2；所有的數(shù)都不會大于100。
Output
對于每組數(shù)據(jù)，輸出兩點(x1,y1),(x2,y2)之間的折線距離。注意輸出結(jié)果精確到小數(shù)點后3位。
Sample Input
5
0 0 0 1
0 0 1 0
2 3 3 1
99 99 9 9
5 5 5 5
Sample Output
1.000
2.414
10.646
54985.047
0.000
甜甜從小就喜歡畫圖畫，最近他買了一支智能畫筆，由于剛剛接觸，所以甜甜只會用它來畫直線，于是他就在平面直角坐標(biāo)系中畫出如下的圖形：
甜甜的好朋友蜜蜜發(fā)現(xiàn)上面的圖還是有點規(guī)則的，于是他問甜甜：在你畫的圖中，我給你兩個點，請你算一算連接兩點的折線長度（即沿折線走的路線長度）吧。
Input
第一個數(shù)是正整數(shù)N（≤100）。代表數(shù)據(jù)的組數(shù)。
每組數(shù)據(jù)由四個非負(fù)整數(shù)組成x1，y1，x2，y2；所有的數(shù)都不會大于100。
Output
對于每組數(shù)據(jù)，輸出兩點(x1,y1),(x2,y2)之間的折線距離。注意輸出結(jié)果精確到小數(shù)點后3位。
Sample Input
5
0 0 0 1
0 0 1 0
2 3 3 1
99 99 9 9
5 5 5 5
Sample Output
1.000
2.414
10.646
54985.047
0.000
題目分析:
簡單的數(shù)學(xué)題, 只需要把2點之間的所有線段加進(jìn)去就可以了 .
代碼如下:
MiYu原創(chuàng), 轉(zhuǎn)帖請注明 : 轉(zhuǎn)載自 ______________白白の屋
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int N, x, x0, x1, y, y0, y1;
cin >> N;
while ( N-- )
{
cin >> x0 >> y0 >> x1 >> y1;
if ( x0 + y0 > x1 + y1 )
{
x0 ^= x1 ^= x0 ^= x1;
y0 ^= y1 ^= y0 ^= y1;
}
x = x0 + y0;
y = x1 + y1;
double line = sqrt ( pow ( x0 - x1 * 1.0, 2 ) + pow ( y0 - y1 * 1.0, 2 ) ); // 2點在同一直線上
double len = sqrt ( 2.0 ) * ( x + x1 - x0 + y * ( y - 1.0 ) / 2.0 - x * ( x + 1.0 ) / 2.0 ); //所有 斜率為1的直線的長度和
for ( int i = x; i < y; ++ i )
{
len += sqrt ( 2.0 * i * i + 2.0 * i + 1 ); //x 與 y 之間的全部斜線(斜率不為1)相加
}
printf( "%.3f\n", x == y ? line : len );
}
return 0;
}
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int N, x, x0, x1, y, y0, y1;
cin >> N;
while ( N-- )
{
cin >> x0 >> y0 >> x1 >> y1;
if ( x0 + y0 > x1 + y1 )
{
x0 ^= x1 ^= x0 ^= x1;
y0 ^= y1 ^= y0 ^= y1;
}
x = x0 + y0;
y = x1 + y1;
double line = sqrt ( pow ( x0 - x1 * 1.0, 2 ) + pow ( y0 - y1 * 1.0, 2 ) ); // 2點在同一直線上
double len = sqrt ( 2.0 ) * ( x + x1 - x0 + y * ( y - 1.0 ) / 2.0 - x * ( x + 1.0 ) / 2.0 ); //所有 斜率為1的直線的長度和
for ( int i = x; i < y; ++ i )
{
len += sqrt ( 2.0 * i * i + 2.0 * i + 1 ); //x 與 y 之間的全部斜線(斜率不為1)相加
}
printf( "%.3f\n", x == y ? line : len );
}
return 0;
}
]]>