Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) 总结

A. Slime Combining

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.

You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes
one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v,
you combine them together to create a slime with value v + 1.

You would like to see what the final state of the row is after you've added all n slimes. Please print the values of the
slimes in the row from left to right.

Input

The first line of the input will contain a single integer, n (1 ≤ n ≤ 100 000).

Output

Output a single line with k integers, where k is
the number of slimes in the row after you've finished the procedure described in the problem statement. The i-th of these numbers should be the value
of the i-th slime from the left.

Sample test(s)

input

1

output

1

input

2

output

2

input

3

output

2 1

input

8

output

4

Note

In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.

In the second sample, we perform the following steps:

Initially we place a single slime in a row by itself. Thus, row is initially 1.

Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with
value 2. Thus, the final state of the board is 2.

In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 21.

In the last sample, the steps look as follows:

122 133 13 23 2 14直接水过吧，没啥好说的

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define LL __int64LL p(LL x,LL y)
{
LL sum=1;
for(LL i=1;i<=y;i++)
{
sum*=x;
}
return sum;
}
LL N(LL x)
{
LL cnt=0;
while(x)
{
x/=2;
cnt++;
}
return cnt;
}
int main()
{
LL n;
while(scanf("%I64d",&n)!=EOF)
{
LL t=n;
while(t)
{
LL cnt=N(t);
printf("%I64d ",cnt);
t=t-p(2,cnt-1);
}
printf("\n");
}
return 0;
}

B. Guess the Permutation

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Bob has a permutation of integers from 1 to n. Denote
this permutation as p. The i-th element
of p will be denoted as pi.
For all pairs of distinct integers i, j between 1 and n,
he wrote the number ai, j = min(pi, pj).
He writes ai, i = 0 for
all integer i from 1 to n.

Bob gave you all the values of ai, j that
he wrote down. Your job is to reconstruct any permutation that could have generated these values. The input will be formed so that it is guaranteed that there is at least one solution that is consistent with the information given.

Input

The first line of the input will contain a single integer n (2 ≤ n ≤ 50).

The next n lines will contain the values of ai, j.
The j-th number on the i-th line will represent ai, j.
The i-th number on the i-th line will be0.
It's guaranteed that ai, j = aj, i and
there is at least one solution consistent with the information given.

Output

Print n space separated integers, which represents a permutation that could have generated these values. If there are multiple possible solutions, print
any of them.

Sample test(s)

input

20 11 0

output

2 1

input

5
0 2 2 1 22 0 4 1 32 4 0 1 31 1 1 0 12 3 3 1 0

output

2 5 4 1 3

Note

In the first case, the answer can be {1, 2} or {2, 1}.

In the second case, another possible answer is {2, 4, 5, 1, 3}.
题意：给你一个数字n，然后1-n数字之间有很多排列，求得一个排列，使得ai, j = min(pi, pj).(a[i][j]为矩阵中的值）

解：我的做法是求矩阵中每一行或列中有多少个不同的数字（0除外），为1的就是mi++（mi初始值为1)，其他的每行个数-1.

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxm=1e2+20;
int p[maxm];
int a[maxm][maxm];
int x[maxm][maxm];
int b[maxm];
int vis[maxm];
int vit[maxm];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
{
memset(vis,0,sizeof(vis));
for(int j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
if(a[i][j]!=0&&!vis[a[i][j]])
{
b[i]++;
vis[a[i][j]]=1;
}
}
}
int t=n;
int mi=1,ma=n;
memset(vit,0,sizeof(vit));
while(t)
{
for(int i=0;i<n;i++)
{
if(b[i]==1&&!vit[i])
{
p[i]=mi++;
vit[i]=1;
}
}
for(int i=0;i<n;i++)
{
b[i]--;
}
t--;
}
for(int i=0;i<n;i++)
{
printf("%d ",p[i]);
}
printf("\n");
}
return 0;
}