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Little Susie, thanks to her older brother, likes to play with cars. Today she decided to set up a tournament between them. The process of a tournament is described in the next paragraph.
There are n toy cars. Each pair collides. The result of a collision can be one of the following: no car turned over, one car turned over, both cars turned over. A car is good if it turned over in no collision. The results of the collisions are determined by an n × n matrix А: there is a number on the intersection of the ?-th row and j-th column that describes the result of the collision of the ?-th and the j-th car:
Susie wants to find all the good cars. She quickly determined which cars are good. Can you cope with the task?
The first line contains integer n (1 ≤ n ≤ 100) — the number of cars.
Each of the next n lines contains n space-separated integers that determine matrix A.
It is guaranteed that on the main diagonal there are - 1, and - 1 doesn‘t appear anywhere else in the matrix.
It is guaranteed that the input is correct, that is, if Aij = 1, then Aji = 2, if Aij = 3, then Aji = 3, and if Aij = 0, then Aji = 0.
Print the number of good cars and in the next line print their space-separated indices in the increasing order.
3
-1 0 0
0 -1 1
0 2 -1
2
1 3
4
-1 3 3 3
3 -1 3 3
3 3 -1 3
3 3 3 -1
0
【题解】
随便搞搞啊= =并不难
我太弱了只会水题= =
1 #include<stdio.h> 2 #include<string.h> 3 using namespace std; 4 bool k[110]; 5 int a[110][110],n; 6 void g(int &xx) { 7 int x=0; 8 int f=1; char ch=getchar(); 9 while(ch<‘0‘ || ch>‘9‘) {if(ch==‘-‘) f=-1; ch=getchar(); } 10 while(ch>=‘0‘&&ch<=‘9‘) {x=(x<<1)+(x<<3)+ch-‘0‘;ch=getchar();} 11 xx=x*f; 12 } 13 int main() { 14 memset(k,1,sizeof(k)); 15 g(n); 16 for (int i=1;i<=n;++i) for (int j=1;j<=n;++j) g(a[i][j]); 17 for (int i=1;i<=n;++i) for (int j=1;j<=n;++j) { 18 if(a[i][j]==-1||a[i][j]==0) continue; 19 if(a[i][j]==1) k[i]=0; 20 else if(a[i][j]==2) k[j]=0; 21 else k[i]=k[j]=0; 22 } 23 int ans=0; 24 for (int i=1;i<=n;++i) 25 ans+=k[i]; 26 printf("%d\n",ans); 27 for (int i=1;i<=n;++i) if(k[i]) printf("%d ",i); 28 return 0; 29 }
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn‘t equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It‘s time for Susie to go to bed, help her find such string p or state that it is impossible.
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
0001
1011
0011
000
111
impossible
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
【题解】两个01串，求一个串异或两个串所获得的1一样多。
1 #include<stdio.h> 2 #include<string> 3 #include<iostream> 4 using namespace std; 5 string s,t; 6 int main() { 7 cin>>s>>t; 8 int l=s.length(); 9 int c=0; 10 for (int i=0;i<l;++i) 11 if(s[i]!=t[i]) c++; 12 if(c&1) { 13 printf("impossible\n"); 14 return 0; 15 } 16 bool flag=0; 17 for (int i=0;i<l;++i) 18 if(s[i]==t[i]) printf("%c",s[i]); 19 else { 20 if(!flag) printf("%c",s[i]); 21 else printf("%c",t[i]); 22 flag^=1; 23 } 24 printf("\n"); 25 return 0; 26 }
Little girl Susie went shopping with her mom and she wondered how to improve service quality.
There are n people in the queue. For each person we know time ti needed to serve him. A person will be disappointed if the time he waits is more than the time needed to serve him. The time a person waits is the total time when all the people who stand in the queue in front of him are served. Susie thought that if we swap some people in the queue, then we can decrease the number of people who are disappointed.
Help Susie find out what is the maximum number of not disappointed people can be achieved by swapping people in the queue.
The first line contains integer n (1 ≤ n ≤ 105).
The next line contains n integers ti (1 ≤ ti ≤ 109), separated by spaces.
Print a single number — the maximum number of not disappointed people in the queue.
5
15 2 1 5 3
4
Value 4 is achieved at such an arrangement, for example: 1, 2, 3, 5, 15. Thus, you can make everything feel not disappointed except for the person with time 5.
【题解】
就是一道简单题。。。
sort后贪心即可= =
1 #include<stdio.h> 2 #include<algorithm> 3 using namespace std; 4 int main() { 5 int n,a[100010]; 6 scanf("%d",&n); 7 for (int i=1;i<=n;++i)scanf("%d",&a[i]); 8 sort(a+1,a+n+1); 9 int s=0,ss=0; 10 for (int i=1;i<=n;++i) 11 if(s<=a[i]) {++ss;s+=a[i];} 12 printf("%d\n",ss); 13 return 0; 14 }
=========================================
就只会这三题TAT，后面有空补完题......
打Vitual Test.....
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原文地址：http://www.cnblogs.com/TonyNeal/p/cfr303d2_1.html
