A. Majestic 10
The movie "Magnificent 7" has become a western classic. Well, this year we have 10 coaches training the UCF programming teams and once you meet them, you’ll realize why they are called the "Majestic 10"! The number 10 is actually special in many different ways. For example, in basketball, they keep track of various statistics (points scored, rebounds, etc.) and if a player has 10+ (10 or more) in a particular stat, they call it a double.
The Problem:
Given three stats for a basketball player, you are to determine how many doubles the player has, i.e., how many of the stats are greater than or equal to 10.
The Input:
The first input line contains a positive integer, n, indicating the number of players. Each of the following n input lines contains three integers (separated by a space and each between 0 and 100, inclusive), providing the three stats for a player.
The Output:
Print each input line as it appears in the input. Then, on the following output line, print a message indicating how many stats are greater than or equal to 10:
print zilch if none of the three stats is greater than or equal to 10,
print double if one of the three stats is greater than or equal to 10,
print double-double if two of the three stats are greater than or equal to 10,
print triple-double if all three stats are greater than or equal to 10.
Leave a blank line after the output for each player.
样例输入
4
5 0 8
30 10 50
20 5 20
5 100 6
样例输出
5 0 8
zilch
30 10 50
triple-double
20 5 20
double-double
5 100 6
double
#include<iostream>
using namespace std;
int main(){
int n,a[1001],sum;
cin>>n;
while(n--){
sum=0;
for(int i=0;i<3;i++){
cin>>a[i];
if(a[i]>=10){
sum++;
}
}
for(int i=0;i<2;i++){
cout<<a[i]<<" ";
}
cout<<a[2]<<endl;
if(sum==0){
cout<<"zilch"<<endl;
}
if(sum==1){
cout<<"double"<<endl;
}
if(sum==2){
cout<<"double-double"<<endl;
}
if(sum==3){
cout<<"triple-double"<<endl;
}
if(n!=0){
cout<<endl;
}
}
}
B. Phoneme Palindromes
A palindrome is a string that reads the same forward and backward, e.g., madam and abba. Since some letters sound the same (e.g., c and k), we define a phoneme palindrome as a string that sounds the same forward and backward, e.g., cak and ckckbbkcck.
The Problem:
Given the letters that sound the same and a string, you are to determine if the string is a phoneme palindrome.
The Input:
The first input line contains a positive integer, n, indicating the number of test cases to process. Each test case starts with an integer, p (1 ≤ p ≤ 13), indicating the count for pairs of letters that sound the same. Each of the following p input lines provides two distinct lowercase letters (starting in column 1 and separated by a space) that sound the same. Assume that no letter appears in more than one pair. The next input line for a test case contains an integer, q (1 ≤ q ≤ 100), indicating the number of strings to test for phoneme palindrome. Each of the following qinput lines provides a string (starting in column 1 and lowercase letters only) of length 1 to 50, inclusive.
The Output:
For each test case, print the header “Test case #n:”, where n indicates the case number starting with 1. Then print each string for that test case followed by a space, followed by a message (YES or NO) indicating whether or not the string is a phoneme palindrome. Leave a blank line after the output for each test case.
样例输入
2 1 c k 6 a cac ck cab kaak ckckkcck 2 a z x s 5 abbbz asxz cx sxxabzxss ks
样例输出
Test case #1: a YES cac YES ck YES cab NO kaak YES ckckkcck YES Test case #2: abbbz YES asxz YES cx NO sxxabzxss YES ks NO
题解：就是在回文数的基础下再多一个条件如果2个字符不相等但音数相同也是可以的（就是循环从左到右如果a[i]=a[n-i-1]就直接满足了然后continue不用再判断是否音数相等，如果不相等就再来一个循环判断音数相等）
代码：
#include<iostream> #include<string> using namespace std; int t,n,n1,num; string w; struct huiwen { char p; char q; } a[10001]; bool ishuiwen(string w){ int len,flag=0; len=w.size(); if(len==1){//字符只有一个的话肯定是音数回文直接return return 1; } else{ for(int i=0;i<len/2;i++){//判断 if(w[i]==w[len-1-i]){ //如果刚好相等就不用判断音数相等了 continue; } if(w[i]!=w[len-1-i]){//不相同的话，如果音数也不相等直接的话直接return不用继续判断了 for(int j=0;j<n;j++){ if((w[i]==a[j].p&&w[len-1-i]==a[j].q)||(w[i]==a[j].q&&w[len-1-i]==a[j].p)){ flag=1; break; } } } if(flag==0){ return 0; } } return 1; } } int main() { cin>>t; num=1; while(t--) { cin>>n; for(int i=0; i<n; i++) { cin>>a[i].p>>a[i].q; } cin>>n1; cout<<"Test case #"<<num<<":"<<endl; num++; for(int i=0; i<n1; i++) { cin>>w; if(ishuiwen(w)==0){ cout<<w<<" "<<"NO"<<endl; } else{ cout<<w<<" "<<"YES"<<endl; } } if(t!=0){ cout<<endl; } } }
C. Don't Break the Ice
Mr. Pennybags is enthralled with his new board game which involves knocking out tiny plasticice blocks from a square grid. Pennybags was never very good at games and would like to practice his breaking strategies without having to reset the board. He has asked Unified Coders For Games (UCF Games) to develop such a tool. Pennybags wants a program to take in the description of a board and a sequence of moves and determine the number of invalid moves.
The Problem:
Given the dimensions (number of rows and columns) of a square game board and a list of moves, determine the number of attempted moves that would knock out an ice block that is no longer in the board. Note that when an ice block is knocked out, other blocks may fall out of the board as well. More specifically, an ice block will fall unless it is in a complete row (the row contains all its ice blocks) or it is in a complete column (the column contains all its ice blocks). Note also that if the fall of block 
results in the fall of block 
, then other blocks may fall as a result of block falling.
The Input:
The first input line contains a positive integer, t, indicating the number of game boards. This is followed by the data for each game. The first input line for each game contains two integers (separated by a space): the dimensions (length and width) of a square game board (between 1 and 50 inclusive) and the number of attempted moves in Pennybags’ strategy (between 1 and 100 inclusive). This is followed by the row and column (separated by a space) for each attempted move, one move per line. Assume that the input values are valid, e.g., the row/column for an attempted move will always be a cell of the game board.
The Output:
For each game board, output "Strategy #b: i" where b is the game board number (starting with 1), and i is the number of invalid moves. Leave a blank line after the output for each game board.
样例输入
3
4 5
1 1
1 2
4 1
4 2
1 1
4 4
1 3
2 4
1 4
4 4
3 3
1 1
2 2
3 3
样例输出
Strategy #1: 2 Strategy #2: 1 Strategy #3: 0
题解：这题一开始没看懂以为不能行列不能超过1 就是1 1 不能直接变成 1 3这样，qwq。其实就是如果你在这一行这一列那么这一行这一列就不能再用了，如果下一个坐标的行和列都被用过了话sum++就可以了。
#include<iostream> using namespace std; int main(){ int t,l,ge,sum,num=1; int x,y; int a[1001],b[1001];//用2个数组代表x轴和y轴 cin>>t; while(t--){ sum=0; cin>>l>>ge; for(int i=1;i<=l;i++){//把每一个坐标设成1代表没有走过 a[i]=1; b[i]=1; } for(int i=1;i<=ge;i++){ cin>>x>>y; if(a[x]==0&&b[y]==0){//判断 sum++; } a[x]=0;//如果走过就变成0 b[y]=0; } cout<<"Strategy #"<<num<<":"<<" "<<sum<<endl; num++; if(t!=0){ cout<<endl; } } return 0; }
D. Wildest Dreams
Secretly, Arup loves Taylor Swift's music. He covers for his addiction by claiming that Anya, his three-year-old daughter, forces him to play the Taylor Swift CD in the car whenever she's in it. The reality is that Arup also listens to the CD even when Anya isn't in the car.
Both of them have peculiar listening habits. Anya typically obsesses over a single song and will request that Arup play that song on infinite repeat while she is in the car. For example, the current song Anya is obsessed with is track 9, Wildest Dreams. So, when Anya enters the car, Arup immediately starts playing track 9 (from the beginning of the track) and this song is played repeatedly. Arup, however, wants to listen to the whole CD and not just one track repeatedly. So, whenever Anya exits the car, Arup just lets the CD play in its natural ordering (tracks play in order and when the end of the CD is reached, track 1 starts again). If Anya exits the car in the middle of her favorite song playing, Arup just lets it continue to play that track until its end and then advances to the next song. For our example, if Anya exits the car in the middle of track 9, Arup lets track 9 continue until its end and then advance to the next song (track 10 or if track 9 is the end of CD, track 1).
The Problem:
Arup is curious exactly how long he has listened to Anya's favorite song. Given a list of the lengths of each song, when Anya is in the car and the song Anya is obsessed with, calculate the amount of time Arup has to hear the song that Anya is obsessed with. Recall that when Anya enters the car, her song is played from the beginning and the song keeps repeating as long as Anya is in the car. You may assume that if Anya gets out of the car in the middle of Anya's song playing, Arup will continue listening to it instead of forwarding the CD to the next song but when Anya’s song is finished the next song will continue. If Anya’s song ends exactly when Anya is leaving the car, the next song will continue, i.e., Anya’s song does not loop back to the beginning.
Since Anya changes which CD she's obsessed with periodically, write a program that can solve multiple instances of the problem.
The Input:
The first line of input will contain a single positive integer, n (1 ≤ n ≤ 50), representing the number of CDs Anya has been obsessed with.
The first line of input for each CD will contain two integers: t (1 ≤ t ≤ 20), the number of tracks on the CD, and k (1 ≤ k ≤ t), the track that Anya is obsessed with. The second line of input for each CD contains t space separated positive integers representing the lengths of each of the t tracks on the CD, in order, in seconds. No CD will be more than 86,400 seconds in length. The third line of input for each CD contains a single positive integer, d (1 ≤ d ≤ 100), the number of days to evaluate. The following d lines contain information about each day. Each of these lines will start with an integer, 
(1 ≤ ≤ 20), the number of segments of driving for day i. This is followed by si positive integers, representing the lengths of each of the driving segments, in seconds. Assume that Anya is in the car for the odd-number segments (first, third, fifth, etc.) The sum of the lengths of each driving segment will never exceed 86,400, the number of seconds in a day.
The Output:
At the beginning of each CD, on a single line, output "CD #c:" where c is the CD number (starting with 1). Then, on the following d lines, where d represents the number of days that CD was played, output the number of seconds Arup listed to Anya's favorite song on the CD, for each day, respectively. Leave a blank line after the output for each CD.
样例输入
2 13 9 212 231 231 235 193 219 207 211 220 247 250 195 270 4 3 1000 900 1000 3 10000 10000 10000 1 2000 2 500 600 2 2 100 200 5 1 70 5 300 277 131 10000 58 2 200 50 2 201 50 2 199 50
样例输出
CD #1: 2100 20780 2000 660 CD #2: 70 7335 200 251 200
题解：这题样例看半天没看懂，一直算不出来为什么等于那个，还是审题不严谨那个naya在奇数时段上来没看到。所以大概就是在奇数段如果这首歌没有刚刚好放完，那只能在下一个时段接着放，然后在循环，她一上车就要切到她喜欢的歌，下车后再循环放.
#include<iostream> #include<cstdio> #include<set> #include<queue> #include<stack> #include<vector> #include<bitset> #include<cstring> #include<string> #include<algorithm> using namespace std; typedef long long ll; const int MAXN=1e5+10; inline int read(){ int x=0,y=1; char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-'){ y=-1; } ch=getchar(); } while(ch>='0'&&ch<='9'){ x=(x<<1)+(x<<3)+(ch^48); ch=getchar(); } return x*y; } int a[110];//代表每首歌曲播放时间 int main(){ int T=read(); for(int t=1;t<=T;t++){ int n=read(),m=read(); int it1=0,it2=0,sum=0; //it1为女儿喜欢歌曲开始时间 //it2为女儿喜欢歌曲结束时间 //sum为cd所有歌曲播放总时间 for(int i=1;i<=n;i++){ a[i]=read(); sum+=a[i]; if(i<m) { it1+=a[i]; } } it2=it1+a[m]; printf("CD #%d: ",t); int p=read(); for(int k=1;k<=p;k++){ int ans=0,it=0; //it表示当前时间 int q=read(); for(int i=1;i<=q;i++){ int cnt=read(); if(cnt==0){ continue; } //当女儿在车上时答案直接加上当前路段时间 if(i%2==1){ ans+=cnt; if(cnt%(it2-it1)==0){ it=it2; } else{ it=it1+cnt%(it2-it1); } } //计算女儿不再车上时所听歌曲时间 else{ ans+=(it2-it1)*(cnt/sum); cnt%=sum; if(cnt<=it2-it){ ans+=cnt; } else{ ans+=it2-it; if(cnt>=sum-it+it1){ ans+=cnt-(sum-it+it1); } } it=(it+cnt)%sum; } } printf("%d ",ans); } printf(" "); } return 0; }