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## 哎，此题用LIST感觉还不是那么水呃，一次A过，STL太赞了。当然，O可没有随随便便写出了就，俺可是用心了的（附代码，供STL初学者参考）

Posted by Pzjay at 2009-09-11 11:34:52 on Problem 3125 and last updated at 2009-09-11 11:38:25
```#include<iostream>
#include<list>
#include<algorithm>
using namespace std;
int main()
{
int m,n,k,i,j;
int v,kk,tmp,sum,summ;
list<int>pzj;
list<int>::iterator ita;
list<int>::iterator itb;
scanf("%d",&m);
while(m--)
{
sum=0;
summ=0;
pzj.clear();
scanf("%d%d",&n,&k);
for(i=0;i<n;i++)
{
scanf("%d",&v);
pzj.push_back(v);
if(i==k)
kk=v;
}
bool cs=0;
k++;
while(find(pzj.begin(),pzj.end(),kk)!=pzj.end())
{
summ++;
cs=0;
tmp=pzj.front();
for(i=9;i>tmp;i--)
if(find(pzj.begin(),pzj.end(),i)!=pzj.end())
{
cs=1;
break;
}
if(cs)
{
pzj.pop_front();
pzj.push_back(tmp);
if(summ==k)//防止数组中重复出现目标数字，要记录需要打印的数字的编号，并不时比较打印的是否是目标数字的标号
{
k=n-sum;
summ=0;
}
}
else
{
pzj.pop_front();
sum++;
if(summ==k)
break;
}
}
printf("%d\n",sum);
}
return false;
}```

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