标签：ber nsis rac other http ict term bae following
There are literally dozens of snooker competitions held each year, and team Jinotega tries to attend them all (for some reason they prefer name "snookah")! When a competition takes place somewhere far from their hometown, Ivan, Artsem and Konstantin take a flight to the contest and back.
Jinotega‘s best friends, team Base have found a list of their itinerary receipts with information about departure and arrival airports. Now they wonder, where is Jinotega now: at home or at some competition far away? They know that:
Please help them to determine Jinotega‘s location!
In the first line of input there is a single integer n: the number of Jinotega‘s flights (1 ≤ n ≤ 100). In the second line there is a string of 3 capital Latin letters: the name of Jinotega‘s home airport. In the next n lines there is flight information, one flight per line, in form "XXX->YYY", where "XXX" is the name of departure airport "YYY" is the name of arrival airport. Exactly one of these airports is Jinotega‘s home airport.
It is guaranteed that flights information is consistent with the knowledge of Jinotega‘s friends, which is described in the main part of the statement.
If Jinotega is now at home, print "home" (without quotes), otherwise print "contest".
4
SVO
SVO->CDG
LHR->SVO
SVO->LHR
CDG->SVO
home
3
SVO
SVO->HKT
HKT->SVO
SVO->RAP
contest
In the first sample Jinotega might first fly from SVO to CDG and back, and then from SVO to LHR and back, so now they should be at home. In the second sample Jinotega must now be at RAP because a flight from RAP back to SVO is not on the list.
题意：给你n张机票 包含机票的起点与终点 判断当前是否在家或在外
题解：水
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <stack> 6 #include <queue> 7 #include <cmath> 8 #include <map> 9 #define ll __int64 10 using namespace std; 11 int n; 12 char a[105]; 13 char b[105],c[105]; 14 int main() 15 { 16 scanf("%d",&n); 17 getchar(); 18 scanf("%s",a); 19 int ans=0; 20 for(int i=1;i<=n;i++) 21 { 22 scanf("%s->%s",c); 23 b[0]=c[0]; 24 b[1]=c[1]; 25 b[2]=c[2]; 26 if(strcmp(a,b)==0) 27 { 28 ans++; 29 } 30 } 31 if(n%2!=0||ans!=(n/2)) 32 { 33 printf("contest\n"); 34 } 35 else 36 printf("home\n"); 37 return 0; 38 }
题意：规定了一种字符串的处理方式 对于遇到的第一个字符用‘a’替换之后出现的相同字符 之后找到下一个未被替换的字符 替换为‘b’ 规则同上
题解：给你一段处理过的字符串 判断是否合法. 直接模拟即可。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <stack> 6 #include <queue> 7 #include <cmath> 8 #include <map> 9 #define ll __int64 10 using namespace std; 11 char a[505]; 12 map<char,int> mp; 13 int main() 14 { 15 scanf("%s",a); 16 int len=strlen(a); 17 char be=‘a‘; 18 mp.clear(); 19 for(int i=0;i<len;i++) 20 { 21 if(mp[a[i]]==1) 22 continue; 23 if(a[i]==be) 24 { 25 be++; 26 } 27 else 28 { 29 printf("NO\n"); 30 return 0; 31 } 32 for(int j=i;j<len;j++) 33 { 34 if(a[i]==a[j]) 35 { 36 mp[a[i]]=1; 37 } 38 } 39 } 40 printf("YES\n"); 41 return 0; 42 }
Misha and Vanya have played several table tennis sets. Each set consists of several serves, each serve is won by one of the players, he receives one point and the loser receives nothing. Once one of the players scores exactly k points, the score is reset and a new set begins.
Across all the sets Misha scored a points in total, and Vanya scored b points. Given this information, determine the maximum number of sets they could have played, or that the situation is impossible.
Note that the game consisted of several complete sets.
The first line contains three space-separated integers k, a and b (1 ≤ k ≤ 109, 0 ≤ a, b ≤ 109, a + b > 0).
If the situation is impossible, print a single number -1. Otherwise, print the maximum possible number of sets.
11 11 5
1
11 2 3
-1
Note that the rules of the game in this problem differ from the real table tennis game, for example, the rule of "balance" (the winning player has to be at least two points ahead to win a set) has no power within the present problem.
题意：两人打乒乓球比赛 双方中只要有一人得k分则本局结束 给你a b分别为两人的总得分 问 最多打了多少局比赛 不可能则输出-1
题解：重点是考虑两人对k取余的结果
若余数不为零 则 对方的获胜场数一定大于零。考虑一下。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <stack> 6 #include <queue> 7 #include <cmath> 8 #include <map> 9 #define ll __int64 10 using namespace std; 11 int main() 12 { 13 ll k,a,b; 14 scanf("%I64d %I64d %I64d",&k,&a,&b); 15 ll ans=0,ansa,ansb; 16 ans=a/k; 17 ansa=ans; 18 a=a%k; 19 ans+=(b/k); 20 ansb=b/k; 21 b=b%k; 22 if((ansb==0&&a!=0)||(ansa==0&&b!=0)) 23 printf("-1\n"); 24 else 25 printf("%I64d\n",ans); 26 return 0; 27 }
Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.
Let [n] denote the set {1, ..., n}. We will also write f: [x] → [y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.
Now then, you are given a function f: [n] → [n]. Your task is to find a positive integer m, and two functions g: [n] → [m], h: [m] → [n], such that g(h(x)) = x for all 
, and h(g(x)) = f(x) for all 
, or determine that finding these is impossible.
The first line contains an integer n (1 ≤ n ≤ 105).
The second line contains n space-separated integers — values f(1), ..., f(n) (1 ≤ f(i) ≤ n).
If there is no answer, print one integer -1.
Otherwise, on the first line print the number m (1 ≤ m ≤ 106). On the second line print n numbers g(1), ..., g(n). On the third line print m numbers h(1), ..., h(m).
If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.
3
1 2 3
3
1 2 3
1 2 3
3
2 2 2
1
1 1 1
2
2
2 1
-1
题意：一个没有想出的构造题 给你长度为n的f(x) 构造 序列 g(n) h(m)
使得 m最小并且g(h(x)) = x , h(g(x)) = f(x)
题解：
g(h(x))= x
h(g(x)) = f(x)
h(g(h(x)))=f(h(x))
得h(x) = f(h(x))
继而得到f(x) = f(f(x))
h(m)为f(x)去重排序
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <stack> 6 #include <queue> 7 #include <cmath> 8 #include <map> 9 #define ll __int64 10 using namespace std; 11 int n; 12 int a[100005]; 13 int g[1000006]; 14 int h[1000006]; 15 int main() 16 { 17 scanf("%d",&n); 18 for(int i=1;i<=n;i++) 19 scanf("%d",&a[i]); 20 int be=1; 21 memset(h,0,sizeof(h)); 22 for(int i=1;i<=n;i++) 23 { 24 if(i==a[i]) 25 { 26 g[be]=i; 27 h[i]=be; 28 be++; 29 } 30 } 31 for(int i=1;i<=n;i++) 32 { 33 if(h[a[i]]==0) 34 { 35 printf("-1\n"); 36 return 0; 37 } 38 } 39 printf("%d\n",be-1); 40 for(int i=1;i<=n;i++) 41 printf("%d ",h[a[i]]); 42 printf("\n"); 43 for(int i=1;i<be;i++) 44 printf("%d ",g[i]); 45 printf("\n"); 46 return 0; 47 }
标签：ber nsis rac other http ict term bae following
原文地址：http://www.cnblogs.com/hsd-/p/6403406.html
