zhejiang省赛热身赛之---周赛第二场
2013-03-18 11:25
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Accurately Say "CocaCola"!
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 25 Accepted Submission(s) : 8
[align=left]Problem Description[/align]
In a party held by CocaCola company, several students stand in a circle and play a game.
One of them is selected as the first, and should say the number 1. Then they continue to count number from 1 one by one (clockwise). The game is interesting in that, once someone counts a number which is a multiple of 7 (e.g. 7, 14, 28, ...) or contains
the digit '7' (e.g. 7, 17, 27, ...), he shall say "CocaCola" instead of the number itself.
For example, 4 students play this game. At some time, the first one says 25, then the second should say 26. The third should say "CocaCola" because 27 contains the digit '7'. The fourth one should say "CocaCola" too, because 28 is a multiple of 7. Then
the first one says 29, and the game goes on. When someone makes a mistake, the game ends.
During a game, you may hear a consecutive of p "CocaCola"s. So what is the minimum number that can make this situation happen?
For example p = 2, that means there are a consecutive of 2 "CocaCola"s. This situation happens in 27-28 as stated above. 27 is then the minimum number to make this situation happen.
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 100) which is the number of test cases. And it will be followed byT consecutive test cases.
There is only one line for each case. The line contains only one integer
p (1 <= p <= 99).
Output
Results should be directed to standard output. The output of each test case should be a single integer in one line, which is the minimum possible number for the first of thep "CocaCola"s stands for.
Sample Input
Sample Output
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
代码:
Build The Electric System
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 3 Accepted Submission(s) : 3
[align=left]Problem Description[/align]
In last winter, there was a big snow storm in South China. The electric system was damaged seriously. Lots of power lines were broken and lots of villages lost contact with the main power grid. The government wants to reconstruct
the electric system as soon as possible. So, as a professional programmer, you are asked to write a program to calculate the minimum cost to reconstruct the power lines to make sure there's at least one way between every two villages.
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 50) which is the number of test cases. And it will be followed byT consecutive test cases.
In each test case, the first line contains two positive integers N andE (2 <=
N <= 500, N <= E <= N * (N - 1) / 2), representing the number of the villages and the number of the original power lines between villages. There followE lines, and each of them contains three integers,
A, B,K (0 <= A, B < N, 0 <=
K < 1000). A and B respectively means the index of the starting village and ending village of the power line. IfK is 0, it means this line still works fine after the snow storm. If
K is a positive integer, it means this line will cost K to reconstruct. There will be at most one line between any two villages, and there will not be any line from one village to itself.
Output
For each test case in the input, there's only one line that contains the minimum cost to recover the electric system to make sure that there's at least one way between every two villages.
Sample Input
Sample Output
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
code:
Easy Task
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 11 Accepted Submission(s) : 7
[align=left]Problem Description[/align]
Calculating the derivation of a polynomial is an easy task. Given a function f(x) , we use (f(x))' to denote its derivation. We use x^n to denote xn. To calculate the derivation
of a polynomial, you should know 3 rules:
(1) (C)'=0 where C is a constant.
(2) (Cx^n)'=C*n*x^(n-1) where n>=1 and C is a constant.
(3) (f1(x)+f2(x))'=(f1(x))'+(f2(x))'.
It is easy to prove that the derivation a polynomial is also a polynomial.
Here comes the problem, given a polynomial f(x) with non-negative coefficients, can you write a program to calculate the derivation of it?
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 1000) which is the number of test cases. And it will be followed byT consecutive test cases.
There are exactly 2 lines in each test case. The first line of each test case is a single line containing an integerN (0 <=
N <= 100). The second line contains N + 1 non-negative integers,CN,
CN-1, ...,C1,
C0, ( 0 <=Ci <= 1000), which are the coefficients of f(x).Ci is the coefficient of the term
with degreei in f(x). (CN!=0)
Output
For each test case calculate the result polynomial g(x) also in a single line.
(1) If g(x) = 0 just output integer 0.otherwise
(2) suppose g(x)= Cmx^m+Cm-1x^(m-1)+...+C0 (Cm!=0),then
output the integers Cm,Cm-1,...C0.
(3) There is a single space between two integers but no spaces after the last integer.
Sample Input
Sample Output
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
Give Me the Number
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 9 Accepted Submission(s) : 2
[align=left]Problem Description[/align]
Numbers in English are written down in the following way (only numbers less than109 are considered). Numberabc,def,ghi is written as "[abc]
million [def] thousand[ghi]". Here "[xyz] " means the written down number
xyz .
In the written down number the part "[abc] million" is omitted if
abc = 0 , "[def] thousand" is omitted if def = 0 , and "[ghi] " is omitted ifghi = 0 . If the whole number is equal to
0 it is written down as "zero". Note that words "million" and "thousand" are singular even if the number of millions or thousands respectively is greater than one.
Numbers under one thousand are written down in the following way. The numberxyz is written as "[x] hundred and[yz] ”. ( If
yz = 0 it should be only “[x] hundred”. Otherwise ify = 0 it should be only “[x] hundred and [z]”.) Here "[x] hundred and" is omitted ifx = 0 . Note that "hundred" is also always singular.
Numbers under 20 are written down as "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", and "nineteen"
respectively. Numbers from 20 to 99 are written down in the following way. Numberxy is written as "[x0] [y] ", and numbers divisible by ten are written as "twenty", "thirty", "forty", "fifty", "sixty", "seventy",
"eighty", and "ninety" respectively.
For example, number 987,654,312 is written down as "nine hundred and eighty seven million six hundred and fifty four thousand three hundred and twelve", number100,000,037 as "one hundred million thirty seven", number
1,000as "one thousand". Note that "one" is never omitted for millions, thousands and hundreds.
Give you the written down words of a number, please give out the original number.
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 1900) which is the number of test cases. It will be followed byT consecutive test cases.
Each test case contains only one line consisting of a sequence of English words representing a number.
Output
For each line of the English words output the corresponding integer in a single line. You can assume that the integer is smaller than109.
Sample Input
Sample Output
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
Faster, Higher, Stronger
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 23 Accepted Submission(s) : 8
[align=left]Problem Description[/align]
In the year 2008, the 29th Olympic Games will be held in Beijing. This will signify the prosperity of China and Beijing Olympics is to be a festival for people all over the world as well.
The motto of Olympic Games is "Citius, Altius, Fortius", which means "Faster, Higher, Stronger".
In this problem, there are some records in the Olympic Games. Your task is to find out which one is faster, which one is higher and which one is stronger.
Input
Standard input will contain multiple test cases. The first line of the input is a single integer
T (1 <= T <= 50) which is the number of test cases. And it will be followed by
T consecutive test cases.
Each test case has 3 lines. The first line is the type of the records, which can only be "Faster" "Higher" or "Stronger". The second line is a positive integer
N meaning the number of the records in this test case. The third line has
N positive integers, i.e. the records data. All the integers in this problem are smaller than 2008.
Output
Results should be directed to standard output. The output of each test case should be a single integer in one line. If the type is "Faster", the records are time records and you should output the fastest one. If the type is "Higher", the records are length
records. You should output the highest one. And if the type is "Stronger", the records are weight records. You should output the strongest one.
Sample Input
Sample Output
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 25 Accepted Submission(s) : 8
[align=left]Problem Description[/align]
In a party held by CocaCola company, several students stand in a circle and play a game.
One of them is selected as the first, and should say the number 1. Then they continue to count number from 1 one by one (clockwise). The game is interesting in that, once someone counts a number which is a multiple of 7 (e.g. 7, 14, 28, ...) or contains
the digit '7' (e.g. 7, 17, 27, ...), he shall say "CocaCola" instead of the number itself.
For example, 4 students play this game. At some time, the first one says 25, then the second should say 26. The third should say "CocaCola" because 27 contains the digit '7'. The fourth one should say "CocaCola" too, because 28 is a multiple of 7. Then
the first one says 29, and the game goes on. When someone makes a mistake, the game ends.
During a game, you may hear a consecutive of p "CocaCola"s. So what is the minimum number that can make this situation happen?
For example p = 2, that means there are a consecutive of 2 "CocaCola"s. This situation happens in 27-28 as stated above. 27 is then the minimum number to make this situation happen.
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 100) which is the number of test cases. And it will be followed byT consecutive test cases.
There is only one line for each case. The line contains only one integer
p (1 <= p <= 99).
Output
Results should be directed to standard output. The output of each test case should be a single integer in one line, which is the minimum possible number for the first of thep "CocaCola"s stands for.
Sample Input
2 2 3
Sample Output
27 70
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
代码:
#include<iostream> #include<vector> #include<queue> #include<stack> #include<stdio.h> #include<algorithm> #include<cstdio> #include<string> #include<cstring> using namespace std;//头文件 int main() { int cas; int n; cin>>cas; int num=0; while(cas--) { cin>>n; if(n==1) num=7; else if(n==2) num=27; else if(n>=3&&n<=10) num=70; else if(n==11) num=270; else num=700; cout<<num<<endl; } return 0; }
Build The Electric System
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 3 Accepted Submission(s) : 3
[align=left]Problem Description[/align]
In last winter, there was a big snow storm in South China. The electric system was damaged seriously. Lots of power lines were broken and lots of villages lost contact with the main power grid. The government wants to reconstruct
the electric system as soon as possible. So, as a professional programmer, you are asked to write a program to calculate the minimum cost to reconstruct the power lines to make sure there's at least one way between every two villages.
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 50) which is the number of test cases. And it will be followed byT consecutive test cases.
In each test case, the first line contains two positive integers N andE (2 <=
N <= 500, N <= E <= N * (N - 1) / 2), representing the number of the villages and the number of the original power lines between villages. There followE lines, and each of them contains three integers,
A, B,K (0 <= A, B < N, 0 <=
K < 1000). A and B respectively means the index of the starting village and ending village of the power line. IfK is 0, it means this line still works fine after the snow storm. If
K is a positive integer, it means this line will cost K to reconstruct. There will be at most one line between any two villages, and there will not be any line from one village to itself.
Output
For each test case in the input, there's only one line that contains the minimum cost to recover the electric system to make sure that there's at least one way between every two villages.
Sample Input
1 3 3 0 1 5 0 2 0 1 2 9
Sample Output
5
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
code:
#include<cstdio> #include<cstring> #include<algorithm> #include<iostream> using namespace std; #define inf 1000111222 #define N 3009 int n,m; ///n--节点数, m--边数 int G ; ///邻接矩阵存储图的信息 int dis ; ///距离 bool used ; ///标记某个点是否已经被使用过 void prim() { int i,j,start=0,ans=0,minD,which; ///认为起点是 1 fill(dis, dis+N, inf); ///所有距离初始化为无穷大 dis[start]=0; ///起点的距离为 0 memset(used, false, sizeof(used)); ///所有点都未被使用过 for(i=0;i < n;i++) /// n 次 { minD = inf; for(j=0;j < n;j++) if(!used[j] && minD >= dis[j]) ///找最小的且没有使用过的点 { minD = dis[j]; which = j; } used[which] = true; ///标记为使用过 ans += dis[which]; /// ans += minD; for(j=0;j < n;j++) if(dis[j] > G[which][j]) ///更新dis值 dis[j] = G[which][j]; } //printf("the min total len is %d\n",ans); printf("%d\n",ans); } int main() { int i,j,st,en,len; int m; int cas; cin>>cas; while(cas--) { cin>>n>>m; for(i=0;i < n;i++) for(j=0;j < n;j++) G[i][j] = inf; ///初始化所有距离为无穷大 while(m--) { scanf("%d %d %d",&st,&en,&len); if(st != en && G[st][en] > len) ///去除重边和自环!!!! G[st][en] = G[en][st] = len; ///无向图 } prim(); } return 0; }
Easy Task
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 11 Accepted Submission(s) : 7
[align=left]Problem Description[/align]
Calculating the derivation of a polynomial is an easy task. Given a function f(x) , we use (f(x))' to denote its derivation. We use x^n to denote xn. To calculate the derivation
of a polynomial, you should know 3 rules:
(1) (C)'=0 where C is a constant.
(2) (Cx^n)'=C*n*x^(n-1) where n>=1 and C is a constant.
(3) (f1(x)+f2(x))'=(f1(x))'+(f2(x))'.
It is easy to prove that the derivation a polynomial is also a polynomial.
Here comes the problem, given a polynomial f(x) with non-negative coefficients, can you write a program to calculate the derivation of it?
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 1000) which is the number of test cases. And it will be followed byT consecutive test cases.
There are exactly 2 lines in each test case. The first line of each test case is a single line containing an integerN (0 <=
N <= 100). The second line contains N + 1 non-negative integers,CN,
CN-1, ...,C1,
C0, ( 0 <=Ci <= 1000), which are the coefficients of f(x).Ci is the coefficient of the term
with degreei in f(x). (CN!=0)
Output
For each test case calculate the result polynomial g(x) also in a single line.
(1) If g(x) = 0 just output integer 0.otherwise
(2) suppose g(x)= Cmx^m+Cm-1x^(m-1)+...+C0 (Cm!=0),then
output the integers Cm,Cm-1,...C0.
(3) There is a single space between two integers but no spaces after the last integer.
Sample Input
3 0 10 2 3 2 1 3 10 0 1 2
Sample Output
0 6 2 30 0 1
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
#include <iostream> #include <cmath> #include <cstring> using namespace std; int xishu[1005]; int op[1005]; int main() { int testcase; cin>>testcase; for(int i=0;i<testcase;i++) { memset(xishu,0,sizeof(xishu)); memset(op,0,sizeof(op)); int cishu,total; cin>>cishu; if(cishu==0) { int a; cin>>a; cout<<"0"<<endl; } else { total=cishu; for(int i=0;i<=cishu;i++) { cin>>xishu[i]; op[i]=xishu[i]*total; total--; } for(int j=0;j<cishu;j++) { if(j!=cishu-1) { cout<<op[j]<<" "; } else { cout<<op[j]<<endl; } } } } return 0; }
Give Me the Number
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 9 Accepted Submission(s) : 2
[align=left]Problem Description[/align]
Numbers in English are written down in the following way (only numbers less than109 are considered). Numberabc,def,ghi is written as "[abc]
million [def] thousand[ghi]". Here "[xyz] " means the written down number
xyz .
In the written down number the part "[abc] million" is omitted if
abc = 0 , "[def] thousand" is omitted if def = 0 , and "[ghi] " is omitted ifghi = 0 . If the whole number is equal to
0 it is written down as "zero". Note that words "million" and "thousand" are singular even if the number of millions or thousands respectively is greater than one.
Numbers under one thousand are written down in the following way. The numberxyz is written as "[x] hundred and[yz] ”. ( If
yz = 0 it should be only “[x] hundred”. Otherwise ify = 0 it should be only “[x] hundred and [z]”.) Here "[x] hundred and" is omitted ifx = 0 . Note that "hundred" is also always singular.
Numbers under 20 are written down as "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", and "nineteen"
respectively. Numbers from 20 to 99 are written down in the following way. Numberxy is written as "[x0] [y] ", and numbers divisible by ten are written as "twenty", "thirty", "forty", "fifty", "sixty", "seventy",
"eighty", and "ninety" respectively.
For example, number 987,654,312 is written down as "nine hundred and eighty seven million six hundred and fifty four thousand three hundred and twelve", number100,000,037 as "one hundred million thirty seven", number
1,000as "one thousand". Note that "one" is never omitted for millions, thousands and hundreds.
Give you the written down words of a number, please give out the original number.
Input
Standard input will contain multiple test cases. The first line of the input is a single integerT (1 <=
T <= 1900) which is the number of test cases. It will be followed byT consecutive test cases.
Each test case contains only one line consisting of a sequence of English words representing a number.
Output
For each line of the English words output the corresponding integer in a single line. You can assume that the integer is smaller than109.
Sample Input
3 one eleven one hundred and two
Sample Output
1 11 102
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
#include<iostream> #include<vector> #include<queue> #include<stack> #include<stdio.h> #include<algorithm> #include<cstdio> #include<string> #include<cstring> #include<map> using namespace std;//头文件 map<string,int> pack; void init() { // pack["hundred"]=100; // pack["million"]=1000000; // pack["thousand"]=1000; pack["one"]=1; pack["two"]=2; pack["three"]=3; pack["four"]=4; pack["five"]=5; pack["six"]=6; pack["seven"]=7; pack["eight"]=8; pack["nine"]=9; pack["ten"]=10; pack["eleven"]=11; pack["twelve"]=12; pack["thirteen"]=13; pack["fourteen"]=14; pack["fifteen"]=15; pack["sixteen"]=16; pack["seventeen"]=17; pack["eighteen"]=18; pack["nineteen"]=19; pack["twenty"]=20; pack["thirty"]=30; pack["forty"]=40; pack["fifty"]=50; pack["sixty"]=60; pack["seventy"]=70; pack["eighty"]=80; pack["ninety"]=90; } int main() { int testcase; cin>>testcase; init(); getchar(); while (testcase--) { map<string,int>::iterator it; string tar[1005]; char s1[1005]; int k=0; int sum=0,tmp=0; int flaghundred=0; int flagmillion=0; int flagthousand=0; int thou=0; int mill=0; cin.getline(s1,1005,'\n'); int len=strlen(s1); for (int i=0;i<len;i++) { if (s1[i]==' '||i==len) { k++; continue; } tar[k]+=s1[i]; } for ( int i=0;i<=k;i++) { if (tar[i]=="and") continue; if ( tar[i]=="hundred") flaghundred=1; else if (tar[i]=="million") flagmillion=1; else if (tar[i]=="thousand") flagthousand=1; for (it=pack.begin();it!=pack.end();it++) { if ((it->first)==tar[i]) { tmp=(it->second); break; } } if (flagmillion==1) { mill=sum; tmp=0; sum=0; flagmillion=0; continue; } if (flagthousand==1) { thou=sum; tmp=0; sum=0; flagthousand=0; continue; } if (flaghundred==1) { tmp=tmp*100; sum=0; flaghundred=0; } sum+=tmp; } if (mill!=0&&thou!=0&&sum!=0) cout<<mill*1000000+1000*thou+sum<<endl; if (mill!=0&&thou!=0&&sum==0) cout<<mill*1000000+1000*thou<<endl; if (mill!=0&&thou==0&&sum!=0) cout<<mill*1000000+sum<<endl; if (mill!=0&&thou==0&&sum==0) cout<<mill*1000000<<endl; if (mill==0&&thou!=0&&sum!=0) cout<<thou*1000+sum<<endl; if (mill==0&&thou!=0&&sum==0) cout<<thou*1000<<endl; if (mill==0&&thou==0) cout<<sum<<endl; mill=0; thou=0; sum=0; } return 0; }
Faster, Higher, Stronger
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 23 Accepted Submission(s) : 8
[align=left]Problem Description[/align]
In the year 2008, the 29th Olympic Games will be held in Beijing. This will signify the prosperity of China and Beijing Olympics is to be a festival for people all over the world as well.
The motto of Olympic Games is "Citius, Altius, Fortius", which means "Faster, Higher, Stronger".
In this problem, there are some records in the Olympic Games. Your task is to find out which one is faster, which one is higher and which one is stronger.
Input
Standard input will contain multiple test cases. The first line of the input is a single integer
T (1 <= T <= 50) which is the number of test cases. And it will be followed by
T consecutive test cases.
Each test case has 3 lines. The first line is the type of the records, which can only be "Faster" "Higher" or "Stronger". The second line is a positive integer
N meaning the number of the records in this test case. The third line has
N positive integers, i.e. the records data. All the integers in this problem are smaller than 2008.
Output
Results should be directed to standard output. The output of each test case should be a single integer in one line. If the type is "Faster", the records are time records and you should output the fastest one. If the type is "Higher", the records are length
records. You should output the highest one. And if the type is "Stronger", the records are weight records. You should output the strongest one.
Sample Input
3 Faster 3 10 11 12 Higher 4 3 5 4 2 Stronger 2 200 200
Sample Output
10
5200
[align=left]Source[/align]
The 5th Zhejiang Provincial Collegiate Programming Contest
#include<iostream> #include<vector> #include<queue> #include<stack> #include<stdio.h> #include<algorithm> #include<cstdio> #include<string> #include<cstring> using namespace std;//头文件 int cmp1(int a,int b) { return a<b; } int cmp2(int a,int b) { return a>b; } int main() { int cas; cin>>cas; char ch[3000]; int flag; int n; int num[3000]; while(cas--) { flag=0; cin>>ch; if(ch[0]=='F') flag=1; else flag=2; cin>>n; for(int i=0;i<n;i++) { cin>>num[i]; } if(flag==1) sort(num,num+n,cmp1); if(flag==2) sort(num,num+n,cmp2); cout<<num[0]<<endl; } return 0; }
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