[简单题]zju_2481
Unique Ascending Array--------------------------------------------------------------------------------
Time limit: 1 Seconds Memory limit: 32768K
Total Submit: 1086 Accepted Submit: 501
--------------------------------------------------------------------------------
Given an array of integers A, you are asked to decide the shortest array of integers B, such that the following two conditions hold.
For all integers 0 <= i < N, there exists an integer 0 <= j < M, such that A == B
For all integers 0 =< i < j < M, we have B < B
Notice that for each array A[] a unique array B[] exists.
Input
The input consists of several test cases. For each test case, an integer N (1 <= N <= 100) is given, followed by N integers A, A, ..., A in a line. A line containing only a zero indicates the end of input.
Output
For each test case in the input, output the array B in one line. There should be exactly one space between the numbers, and there should be no initial or trailing spaces.
Sample Input
8 1 2 3 4 5 6 7 8
8 8 7 6 5 4 3 2 1
8 1 3 2 3 1 2 3 1
0
Sample Output
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
1 2 3
Author: SHI, Xiaohan
很简单，不过我不过，会不会数据很大呢？ int search(int array[],int number,int post){
for(int i=post;i>=0;i--){
if(number<array) continue;
else if(number==array) {
return -1;
}
else if(number>array) {
return i+1;
}
}
return 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
int count; // Size of A
while(scanf(\"%d\",&count),count!=0){
int b;
int bi=-1;
int location;
int element;
for(int i=0;i<count;i++){
scanf(\"%d\",&element); //读入一个元素
int location=search(b,element,bi);
if(location==-1)continue;
for(int j=bi+1;j>=location;j--){
b=b;
}
b=element;
bi++;
}
b=-1;
int index=0;
while(b!=-1){
printf(\"%d \",b);
index++;
}
printf(\"\\n\");
}
} #include <iostream>
#include <fstream>
using namespace std;
int main()
{
bool appear;
ifstream cin(\"in.txt\");
ofstream cout(\"out.txt\");
unsigned long times,i,k,j,max;
while(1)
{
cin >> times;
max = 0;
for(k = 0;k< 65536;k++)
{
appear = false;
}
if(times == 0)
{
break;
}
else
{
for(j = 0;j < times;j++)
{
cin >> i;
appear = true;
if(i > max)
{
max = i;
}
}
}
for(k = 0;k<=max;k++)
{
if(k == max)
{
cout <<k<<endl;
}
else if(appear)
{
cout <<k<<\" \";
}
}
}
}
我就真的不知道错哪里了。。。。 这道题的意思说,在一个序列中,找出满足以下条件的序列B
序列B 的数字都要在A中出现.
序列B中的数字由小到大
可见,序列B是序列A中的数字的一个排序,但是不输出相同的数字 A了
#include<stdio.h>
int main()
{
int shu,i,j,k,n;
freopen(\"in.txt\",\"r\",stdin);
while(scanf(\"%d\",&n),n)
{
for (i=0;i<n;i++) scanf(\"%d\",&shu);
for (i=0;i<n;i++)//排序
for (j=i+1;j<n;j++)
if(shu<shu)
{
k=shu;
shu=shu;
shu=k;
}
printf(\"%d\",shu);
for (i=1;i<n;i++) if(shu>shu)
printf(\" %d\",shu);
printf(\"\\n\");
}
return 0;
} 小康的代码怎么比大象的少那么多？？？晕！！！！ 不知道这样对不对?
#include<stdio.h>
main()
{
int a,b,i,j,k,n;
while(scanf(\"%d\",&n),n)
{
for(i=0;i<n;i++)
scanf(\"%d\",&a);
for(i=0;i<n;i++)
{for(j=i+1;j<n;j++)
if(a<a)
{
k=a;
a=a;
a=k;
}
b=a;
}
printf(\"%d\",b);
for(i=1;i<n;i++)
{
if(b==b)continue;
printf(\" %d\",b);
}
printf(\"\\n\");
}
} 两个程序基本都一样 你仔细看清楚了 你仔细看清楚了,老大 你个数组B好像不用也行吧,还是觉得思路程序都一样 嘻…………但它要求是有B数组的，不用就不符合题目要求了
小康是不是师兄来的？ 你误解题目意思了.数组B,只是题目为了说明意思的.实际编程,你只要输出的答案完全正确就行了.程序要求内存越少,速度越快,答案对,不需要做出错出来.这样就AC 有道理,不愧是高手来的 你有兴趣加到做题群来一起做题
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