You can Solve a Geometry Problem too
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3077 Accepted Submission(s): 1452
Problem Description
Many geometry（幾何）problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is very easy, after all we are now attending an exam, not a contest :)
Give you N (1<=N<=100) segments（線段）, please output the number of all intersections（交點）. You should count repeatedly if M (M>2) segments intersect at the same point.
Note:
You can assume that two segments would not intersect at more than one point.
Input
Input contains multiple test cases. Each test case contains a integer N (1=N<=100) in a line first, and then N lines follow. Each line describes one segment with four float values x1, y1, x2, y2 which are coordinates of the segment’s ending.
A test case starting with 0 terminates the input and this test case is not to be processed.
Output
For each case, print the number of intersections, and one line one case.
Sample Input
2 0.00 0.00 1.00 1.00 0.00 1.00 1.00 0.00 3 0.00 0.00 1.00 1.00 0.00 1.00 1.00 0.000 0.00 0.00 1.00 0.00 0
Sample Output
Author
lcy
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
struct Line
{
double x1,y1,x2,y2;
}node[110];
bool solve(Line a,Line b)
{
if(((a.x1-b.x1)*(a.y2-b.y1)-(a.x2-b.x1)*(a.y1-b.y1))*((a.x1-b.x2)*(a.y2-b.y2)-(a.x2-b.x2)*(a.y1-b.y2))>0)return false;
if(((b.x1-a.x1)*(b.y2-a.y1)-(b.x2-a.x1)*(b.y1-a.y1))*((b.x1-a.x2)*(b.y2-a.y2)-(b.x2-a.x2)*(b.y1-a.y2))>0)return false;
return true;
}
int main()
{
int n;
while(scanf("%d",&n),n)
{
for(int i=0;i<n;i++)scanf("%lf%lf%lf%lf",&node[i].x1,&node[i].y1,&node[i].x2,&node[i].y2);
int res=0;
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if(solve(node[i],node[j]))res++;
}
}
printf("%d\n",res);
}
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/11/01/2230923.html
Write a simple HTML Browser
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3398 Accepted Submission(s): 890
Problem Description
If you ever tried to read a html document on a Macintosh, you know how hard it is if no Netscape is installed.
Now, who can forget to install a HTML browser? This is very easy because most of the times you don't need one on a MAC because there is a Acrobate Reader which is native to MAC. But if you ever need one, what do you do?
Your task is to write a small html-browser. It should only display the content of the input-file and knows only the html commands (tags) <br> which is a linebreak and <hr> which is a horizontal ruler. Then you should treat all tabulators, spaces and newlines as one space and display the resulting text with no more than 80 characters on a line.
Input
The input consists of a text you should display. This text consists of words and HTML tags separated by one or more spaces, tabulators or newlines.
A word is a sequence of letters, numbers and punctuation. For example, "abc,123" is one word, but "abc, 123" are two words, namely "abc," and "123". A word is always shorter than 81 characters and does not contain any '<' or '>'. All HTML tags are either <br> or <hr>.
Output
You should display the the resulting text using this rules:
. If you read a word in the input and the resulting line does not get longer than 80 chars, print it, else print it on a new line.
. If you read a <br> in the input, start a new line.
. If you read a <hr> in the input, start a new line unless you already are at the beginning of a line, display 80 characters of '-' and start a new line (again).
The last line is ended by a newline character.
Sample Input
Hallo, dies ist eine ziemlich lange Zeile, die in Html aber nicht umgebrochen wird. <br> Zwei <br> <br> produzieren zwei Newlines. Es gibt auch noch das tag <hr> was einen Trenner darstellt. Zwei <hr> <hr> produzieren zwei Horizontal Rulers. Achtung mehrere Leerzeichen irritieren Html genauso wenig wie mehrere Leerzeilen.
Sample Output
Hallo, dies ist eine ziemlich lange Zeile, die in Html aber nicht umgebrochen wird. Zwei produzieren zwei Newlines. Es gibt auch noch das tag -------------------------------------------------------------------------------- was einen Trenner darstellt. Zwei -------------------------------------------------------------------------------- -------------------------------------------------------------------------------- produzieren zwei Horizontal Rulers. Achtung mehrere Leerzeichen irritieren Html genauso wenig wie mehrere Leerzeilen.
Source
Recommend
JGShining
刷題練手感！！！！水題啊！
#include<stdio.h>
#include<string.h>
int main()
{
//freopen("test.in","r",stdin);
//freopen("test.out","w",stdout);
char str[100];
int cnt=0;//數每行的字符數
bool first=true;//記錄是否是一行的開頭，開頭不要空格的
while(scanf("%s",&str)!=EOF)
{
if(strcmp(str,"<br>")==0){printf("\n");cnt=0;first=true;continue;}
if(strcmp(str,"<hr>")==0)
{
if(!first){printf("\n");first=true;}
for(int i=0;i<80;i++)printf("-");
printf("\n");first=true;cnt=0;
continue;
}
if(first)
{
printf("%s",str);
cnt+=strlen(str);first=false;
continue;
}
cnt+=(strlen(str)+1);first=false;
if(cnt<=80)
{
printf(" %s",str);
continue;
}
else
{
printf("\n");
printf("%s",str);
cnt=strlen(str);first=false;
}
}
if(!first)printf("\n");//最后以新行結束
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/11/01/2230917.html
Lottery
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1108 Accepted Submission(s): 544
Problem Description
Eddy's company publishes a kind of lottery.This set of lottery which are numbered 1 to n, and a set of one of each is required for a prize .With one number per lottery, how many lottery on average are required to make a complete set of n coupons?
Input
Input consists of a sequence of lines each containing a single positive integer n, 1<=n<=22, giving the size of the set of coupons.
Output
For each input line, output the average number of lottery required to collect the complete set of n coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of ouput.
Sample Input
Sample Output
3 5 11 -- 12 340463 58 ------ 720720
Author
eddy
Recommend
JGShining
題目沒有看懂啊~~~~
/*****************************************************************
HDU 1099
*****************************************************************/
#include<iostream>
using namespace std;
long long gcd(long long a,long long b)
{
if(b==0)return a;
else return gcd(b,a%b);
}
int main()
{
int i,n;
long long number,n1,n2;
long long integer;
long long fenzi,fenmu;
long long a[23];
a[1]=1;
for(i=2;i<=22;i++)a[i]=i*a[i-1]/gcd(i,a[i-1]);
while(scanf("%d",&n)!=EOF)
{
fenzi=0;
for(i=1;i<=n;i++) fenzi+=a[n]/i;
fenzi*=n;
number=gcd(fenzi,a[n]);
fenzi=fenzi/number;
fenmu=a[n]/number;
integer=fenzi/fenmu;
fenzi-=integer*fenmu;
if(fenzi==0)
{
printf("%I64d\n",integer);continue;
}
int size1=0,size2=0;
n1=integer;n2=fenmu;
while(n1!=0){size1++;n1/=10;}
while(n2!=0){size2++;n2/=10;}
for(i=0;i<=size1;i++)printf("");
printf("%I64d\n",fenzi);
printf("%I64d ",integer);
for(i=0;i<size2;i++)printf("-");
printf("\n");
for(i=0;i<=size1;i++)printf("");
printf("%I64d\n",fenmu);
}
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/30/2229004.html
維基百科資料:
卡塔蘭數
卡塔蘭數是組合數學中一個常出現在各種計數問題中出現的數列。由以比利時的數學家歐仁·查理·卡塔蘭 (1814–1894)命名。
卡塔蘭數的一般項公式為  另類遞歸式： h(n)=((4*n-2)/(n+1))*h(n-1);
前幾項為 （OEIS中的數列A000108）: 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, ...
Cn的另一個表達形式為 所以，Cn是一個自然數；這一點在先前的通項公式中并不顯而易見。這個表達形式也是André對前一公式證明的基礎。(見下文的第二個證明。)
卡塔蘭數滿足以下遞推關系
它也滿足
這提供了一個更快速的方法來計算卡塔蘭數。
卡塔蘭數的漸近增長為
它的含義是左式除以右式的商趨向于1當n → ∞。（這可以用n!的斯特靈公式來證明。）
所有的奇卡塔蘭數Cn都滿足n = 2k − 1。所有其他的卡塔蘭數都是偶數。
組合數學中有非常多.的組合結構可以用卡塔蘭數來計數。在Richard P. Stanley的Enumerative Combinatorics: Volume 2一書的習題中包括了66個相異的可由卡塔蘭數表達的組合結構。以下用Cn=3和Cn=4舉若干例：
- Cn表示長度2n的dyck word的個數。Dyck word是一個有n個X和n個Y組成的字串，且所有的部分字串皆滿足X的個數大于等于Y的個數。以下為長度為6的dyck words:
XXXYYY XYXXYY XYXYXY XXYYXY XXYXYY
- 將上例的X換成左括號，Y換成右括號，Cn表示所有包含n組括號的合法運算式的個數:
((())) ()(()) ()()() (())() (()())
- Cn表示所有不同構的含n個分枝結點的滿二叉樹的個數。（一個有根二叉樹是滿的當且僅當每個結點都有兩個子樹或沒有子樹。）
證明：
令1表示進棧，0表示出棧，則可轉化為求一個2n位、含n個1、n個0的二進制數，滿足從左往右掃描到任意一位時，經過的0數不多于1數。顯然含n個1、n個0的2n位二進制數共有 個，下面考慮不滿足要求的數目.
考慮一個含n個1、n個0的2n位二進制數，掃描到第2m+1位上時有m+1個0和m個1（容易證明一定存在這樣的情況），則后面的0-1排列中必有n-m個1和n-m-1個0。將2m+2及其以后的部分0變成1、1變成0，則對應一個n+1個0和n-1個1的二進制數。反之亦然（相似的思路證明兩者一一對應）。
從而 。證畢。
- Cn表示所有在n × n格點中不越過對角線的單調路徑的個數。一個單調路徑從格點左下角出發，在格點右上角結束，每一步均為向上或向右。計算這種路徑的個數等價于計算Dyck word的個數： X代表“向右”，Y代表“向上”。下圖為n = 4的情況：
-
- Cn表示通過連結頂點而將n + 2邊的凸多邊形分成三角形的方法個數。下圖中為n = 4的情況：
- Cn表示對{1, ..., n}依序進出棧的置換個數。一個置換w是依序進出棧的當S(w) = (1, ..., n), 其中S(w)遞歸定義如下：令w = unv，其中n為w的最大元素，u和v為更短的數列；再令S(w) =S(u)S(v)n，其中S為所有含一個元素的數列的單位元。
- Cn表示用n個長方形填充一個高度為n的階梯狀圖形的方法個數。下圖為 n = 4的情況:
百度百科資料:
簡介
中文:卡特蘭數
Catalan數是組合數學中一個常出現在各種計數問題中出現的數列。由以比利時的數學家歐仁·查理·卡塔蘭 (1814–1894)命名。
原理：
令h(0)=1,h(1)＝1,catalan數滿足遞歸式：
h(n)= h(0)*h(n-1) + h(1)*h(n-2) +  + h(n-1)h(0) (其中n>=2)
該遞推關系的解為：
h(n)=C(2n,n)/(n + 1) (n=1,2,3,)
另類遞歸式： h(n)=((4*n-2)/(n+1))*h(n-1);
前幾項為 （OEIS中的數列A000108）: 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452,
應用
我總結了一下，最典型的四類應用：（實質上卻都一樣，無非是遞歸等式的應用，就看你能不能分解問題寫出遞歸式了）
1.括號化問題。
矩陣鏈乘： P=a1×a2×a3×……×an，依據乘法結合律，不改變其順序，只用括號表示成對的乘積，試問有幾種括號化的方案？(h(n)種)
2.出棧次序問題。
一個棧(無窮大)的進棧序列為1,2,3,..n,有多少個不同的出棧序列?
類似：
(1)有2n個人排成一行進入劇場。入場費5元。其中只有n個人有一張5元鈔票，另外n人只有10元鈔票，劇院無其它鈔票，問有多少中方法使得只要有10元的人買票，售票處就有5元的鈔票找零？(將持5元者到達視作將5元入棧，持10元者到達視作使棧中某5元出棧)
(2)在圓上選擇2n個點,將這些點成對連接起來，使得所得到的n條線段不相交的方法數。
3.將多邊行劃分為三角形問題。
將一個凸多邊形區域分成三角形區域的方法數?
類似：一位大城市的律師在她住所以北n個街區和以東n個街區處工作。每天她走2n個街區去上班。如果她
從不穿越（但可以碰到）從家到辦公室的對角線，那么有多少條可能的道路？
類似：在圓上選擇2n個點,將這些點成對連接起來使得所得到的n條線段不相交的方法數?
4.給頂節點組成二叉樹的問題。
給定N個節點，能構成多少種形狀不同的二叉樹？
(一定是二叉樹!
先去一個點作為頂點,然后左邊依次可以取0至N-1個相對應的,右邊是N-1到0個,兩兩配對相乘,就是h(0)*h(n-1) + h(2)*h(n-2) + + h(n-1)h(0)=h(n))
（能構成h（N）個）
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/13/2210935.html
Random Sequence
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 202 Accepted Submission(s): 124
Problem Description
There is a random sequence L whose element are all random numbers either -1 or 1 with the same possibility. Now we define MAVS, the abbreviate of Maximum Absolute Value Subsequence, to be any (if more than one) subsequences of L whose absolute value is maximum among all subsequences. Given the length of L, your task is to find the expectation of the absolute value of MAVS.
Input
There is only one input file. The first line is the number of test cases T. T positive integers follow, each of which contains one positive number not greater than 1500 denoted the length of L.
Output
For each test case, output the expectation you are required to calculate. Answers are rounded to 6 numbers after the decimal point.(as shown in the sample output)
Sample Input
Sample Output
Case 1: 1.000000 Case 2: 2.750000 Case 3: 4.167969
Source
Recommend
lcy
數論神題啊！！！！
卡特蘭數
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
using namespace std;
long double a[2000];
long double p;
long double C(int n){
long double t=1;
for(int i=n+1;i<=2*n+1;i++){
t*=i;
t/=i-n;
}
// for(int i=1;i<=n+1;i++) t/=i;
return t;
}
int main(){
int t,n,cnt=0;
a[1]=1;
for(int i=2;i<=1500;i++){
if(i%2==0) a[i]=(a[i-1]*2+C(i/2-1));
else a[i]=(a[i-1]*2+C(i/2-1)*2);
//cout<<"i="<<i<<","<<a[i]<<endl;
}
cin>>t;
while(t--){
cin>>n;
p=pow(2.0,n-1);
cout<<"Case "<<++cnt<<": "<<fixed<<setprecision(6)<<a[n]/p<<endl;
}
// system("pause");
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/13/2210930.html
A Card Game
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4 Accepted Submission(s): 3
Problem Description
There are N cards on the table, whose front side is writen one integer number from 1 to M. We call one card "a type k card" if its number is k. The quantity of type i cards is a_i.
Let's play a game with these cards. We divide these cards into M piles by random with the only constrains that the quantity of cards in i-th (indexed from 1) pile must exactly be a_i. The possbility of each card appears in i-th pile is directly proportional to the size of this pile. That is to say, if the size of a pile is A, the possibility for each card appears in this pile is A/N assuming that N is the amount of all cards. We choose pile 1 to start the game. Assuming the we now play this game at pile k, we randomly choose a card from pile k with the same possibility for all cards in it, remember the number written on this card and throw it away. If the number on the chosen card is j, we continue this game at pile j on next round. The game terminates when we are going to get a card from an empty pile.
Now the question is, when the game ends, what is the possibility that all piles are empty?
Input
There is only one input file. The first line is the number of test cases T. T cases follow, each of which contains two lines. The first line is an integer M (1 <= M <= 100), the number of type of cards (and the number of piles, they are exactly the same). The second line contains M positive integers not greater than 1000, the i-th number of which is a_i.
Output
For each test case, output the possibility you are required to calculate. Answers are rounded to 6 numbers after the decimal point.(as shown in the sample output)
Sample Input
Sample Output
Case 1: 1.000000 Case 2: 0.333333
水題，看代碼：
這樣的題目很爽！
#include<stdio.h>
int main()
{
int T;
int iCase=0;
int a;
int n;
int ans,res;
scanf("%d",&T);
while(T--)
{
iCase++;
scanf("%d",&n);
scanf("%d",&ans);
res=ans;
for(int i=1;i<n;i++)
{
scanf("%d",&a);
res+=a;
}
printf("Case %d: %.6lf\n",iCase,(double)ans/res);
}
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/07/2200549.html
Working in Beijing
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 659 Accepted Submission(s): 333
Problem Description
Mr. M is an undergraduate student of FDU. He finds an intern position in Beijing, so that he cannot attend all the college activities. But in some conditions, he must come back to Shanghai on certain date. We can assume the important activities that Mr. M must attend are occupy a whole day. Mr. M must take flight to Shanghai before that day and leave after that day. On the other hand, Mr. M is absent in Beijing and he will lose his salary for his absent.
Sometimes the cost of flight is much higher than the loss of salary, so to save the cost on the travel, Mr. M can stay in Shanghai to wait for another important date before he back to Beijing.
Now, Mr. M knows all of the important date in the next year. Help him schedule his travel to optimize the cost.
Input
The input contains several test cases. The first line of single integer indicates the number of test cases.
For each test case, the first line contains three integers: n, a and b, denoting the number of important events, the cost of a single flight from Beijing to Shanghai or Shanghai to Beijing and the salary for a single day stay in Beijing. (1 <= n <= 100000, 1 <= a <= 1000000000, 1 <= b <=100)
Next line contains n integers ti, denoting the time of the important events. You can assume the ti are in increasing order and they are different from each other. (0 <= ti <= 10000000)
Output
For each test case, output a single integer indicating the minimum cost for this year.
Sample Input
2 1 10 10 5 5 10 2 5 10 15 65 70
Sample Output
Source
Recommend
lcy
#include<stdio.h>
int main()
{
long long res;
int n,a,b;
int t1,t2;
int i,T;
int iCase=0;
scanf("%d",&T);
while(T--)
{
iCase++;
scanf("%d%d%d",&n,&a,&b);
res=a+b;
scanf("%d",&t1);
for(i=1;i<n;i++)
{
scanf("%d",&t2);
if((t2-t1-1)*b>2*a) res+=2*a+b;
else res+=b*(t2-t1);
t1=t2;
}
res+=a;
printf("Case #%d: %I64d\n",iCase,res);
}
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/04/2198939.html
Parsing URL
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 520 Accepted Submission(s): 321
Problem Description
In computing, a Uniform Resource Locator or Universal Resource Locator (URL) is a character string that specifies where a known resource is available on the Internet and the mechanism for retrieving it.
The syntax of a typical URL is:
scheme://domain:port/path?query_string#fragment_id
In this problem, the scheme, domain is required by all URL and other components are optional. That is, for example, the following are all correct urls:
http://dict.bing.com.cn/#%E5%B0%8F%E6%95%B0%E7%82%B9
http://www.mariowiki.com/Mushroom
https://mail.google.com/mail/?shva=1#inbox
http://en.wikipedia.org/wiki/Bowser_(character)
ftp://fs.fudan.edu.cn/
telnet://bbs.fudan.edu.cn/
http://mail.bashu.cn:8080/BsOnline/
Your task is to find the domain for all given URLs.
Input
There are multiple test cases in this problem. The first line of input contains a single integer denoting the number of test cases.
For each of test case, there is only one line contains a valid URL.
Output
For each test case, you should output the domain of the given URL.
Sample Input
3 http://dict.bing.com.cn/#%E5%B0%8F%E6%95%B0%E7%82%B9 http://www.mariowiki.com/Mushroom https://mail.google.com/mail/?shva=1#inbox
Sample Output
Case #1: dict.bing.com.cn Case #2: www.mariowiki.com Case #3: mail.google.com
Source
Recommend
lcy
#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
char str[5000];
char tt[5000];
int main()
{
int T;
int iCase=0;
bool start;
int i;
scanf("%d",&T);
while(T--)
{
iCase++;
scanf("%s",&str);
start=false;
int len=strlen(str);
int t=0;
for(i=2;i<len;i++)
{
if(start==false&&str[i-2]==':'&&str[i-1]=='/'&&str[i]=='/') {start=true;continue;}
if(start&&(str[i]=='/'||str[i]==':')) break;
if(start) tt[t++]=str[i];
}
tt[t]='\0';
printf("Case #%d: %s\n",iCase,tt);
}
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/04/2198937.html
Mario and Mushrooms
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 265 Accepted Submission(s): 220
Problem Description
Mario usually relaxes himself by walking along the shady track near the Mushroom Kingdom. The evil King Koopa noticed that and placed a lot of mushroom on the road. There are two types of mushrooms, max mushrooms and bad mushrooms. The bad mushrooms will decrease Mario's HP by m points, on the other hand, max mushrooms will increase Mario's HP by one point. The mushrooms are randomly placed on the track and Mario will receive them one by one. Once Mario's HP becomes zero or below after he received a mushroom, he will die.
Notice that Mario begins with HP zero, so if the first mushroom is bad, Mario will die immediately. Fortunately, if Mario receives all the mushrooms, he will be alive with HP 1. In the other words, if there are k bad mushrooms on the way, there will also be m*k+1 max mushrooms.
Princess Peach wants to know the possibility for Mario staying alive. Please help her to calculate it out.
Input
There are several test cases. The first line contains only one integer T, denoting the number of test cases.
For each test case, there is only one line including two integers: m and k, denoting the amount of points of HP the Mario will decrease if he receives a bad mushroom, and the number of bad mushrooms on the track. (1 <= m <= 1000, 1 <= k <= 1000)
Output
For each test case, output only real number denoting the possibility that Mario will survive if he receives all the randomly placed mushrooms one by one. The answer should be rounded to eight digits after the decimal point.
Sample Input
Sample Output
Case #1: 0.33333333 Case #2: 0.00020488
Source
Recommend
lcy
明顯的概率題。。。。
結果通過樣例猜想就得出來了。。。很簡單的結果，秒過
#include<stdio.h>
int main()
{
int T;
int m,k;
int iCase=0;
scanf("%d",&T);
while(T--)
{
iCase++;
scanf("%d%d",&m,&k);
printf("Case #%d: %.8lf\n",iCase,(double)1/(k+m*k+1));
}
return 0;
}
文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/04/2198936.html
接觸ACM以來的第一場正式比賽，也許會是最后一場比賽了，ACM，短暫的時間通過ACM也學到了不少東西，認識到了不少人，哈哈~~~~~ 文章來源: http://www.cnblogs.com/kuangbin/archive/2011/10/04/2198904.html
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