hdu--6168--Numbers
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Numbers
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 162 Accepted Submission(s): 86
Problem Description
zk has n numbers a1,a2,...,an . For each (i,j) satisfying 1≤i<j≤n, zk generates a new number (ai+aj) . These new numbers could make up a new sequence b1,b2,...,bn(n−1)/2 .
LsF wants to make some trouble. While zk is sleeping, Lsf mixed up sequence a and b with random order so that zk can't figure out which numbers were in a or b. "I'm angry!", says zk.
Can you help zk find out which n numbers were originally in a?
LsF wants to make some trouble. While zk is sleeping, Lsf mixed up sequence a and b with random order so that zk can't figure out which numbers were in a or b. "I'm angry!", says zk.
Can you help zk find out which n numbers were originally in a?
Input
Multiple test cases(not exceed 10).
For each test case:
∙ The first line is an integer m(0≤m≤125250), indicating the total length of a and b. It's guaranteed m can be formed as n(n+1)/2.
∙ The second line contains m numbers, indicating the mixed sequence of a and b.
Eachai is in [1,10^9]
For each test case:
Each
Output
For each test case, output two lines.
The first line is an integer n, indicating the length of sequence a;
The second line should contain n space-seprated integersa1,a2,...,an(a1≤a2≤...≤an) . These are numbers in sequence a.
It's guaranteed that there is only one solution for each case.
The first line is an integer n, indicating the length of sequence a;
The second line should contain n space-seprated integers
It's guaranteed that there is only one solution for each case.
Sample Input
62 2 2 4 4 4211 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11
Sample Output
32 2 261 2 3 4 5 6
题意:
有n个数的序列a(a1......an),可以两两组合成n(n-1)/2个数的b序列。把它们混成一个数组,让你求出原n个数的a序列。
解题思路:
因为b序列是由两个a序列的元素组合而成,所以 混合的这个序列按从小到大排序后,前两个较小的数一定为a序列中的元素。假设为z[1],z[2],那么z[1]+z[2]就一定存在于b数组中。把z[1]+z[2]用map“标记”一下,继续从z[3]遍历混合的数组,判断z[3]有没有被“标记”,如果有 说明这个数为b序列中的数,跳过 继续遍历下一个。如果没有,该数就为a数组中的元素。以此类推.........
代码:
C++ Code
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#include <bits/stdc++.h>
using namespace std;
const int N = 125250 + 8;
map<int, int> m;
int a[N], z[N];
int main()
{
int n, k;
while(~scanf("%d", &n))
{
k = 0;
m.clear();
for(int i = 1; i <= n; i++)
scanf("%d", &z[i]);
sort(z + 1, z + 1 + n);
a[k++] = z[1];
a[k++] = z[2];
m[z[1] + z[2]]++; ///表示b数组中包含该数
for(int i = 3; i <= n; i++)
{
if(m[z[i]] > 0)
{
///如果b数组中有这个数,
m[z[i]]--;
continue;
}
else
{
a[k++] = z[i];
for(int j = 0; j < k - 1; j++)
m[a[j] + z[i]]++;
}
}
sort(a, a + k);
printf("%d\n", k);
for(int i = 0; i < k; i++)
{
if(i)
printf(" ");
printf("%d", a[i]);
}
printf("\n");
}
}
using namespace std;
const int N = 125250 + 8;
map<int, int> m;
int a[N], z[N];
int main()
{
int n, k;
while(~scanf("%d", &n))
{
k = 0;
m.clear();
for(int i = 1; i <= n; i++)
scanf("%d", &z[i]);
sort(z + 1, z + 1 + n);
a[k++] = z[1];
a[k++] = z[2];
m[z[1] + z[2]]++; ///表示b数组中包含该数
for(int i = 3; i <= n; i++)
{
if(m[z[i]] > 0)
{
///如果b数组中有这个数,
m[z[i]]--;
continue;
}
else
{
a[k++] = z[i];
for(int j = 0; j < k - 1; j++)
m[a[j] + z[i]]++;
}
}
sort(a, a + k);
printf("%d\n", k);
for(int i = 0; i < k; i++)
{
if(i)
printf(" ");
printf("%d", a[i]);
}
printf("\n");
}
}
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