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用快排超时,,求解#include<iostream>
using namespace std;
int data[100000]={0};
void swap(int &a,int &b)
{
int temp;
temp=a;
a=b;
b=temp;
}
int partion(int i,int j)
{
int mark=i,x=data[i];
for(int k=i+1;k<=j;k++)
{
if(data[k]<x)
{
mark++;
if(mark!=k)swap(data[mark],data[k]);
}
}
if(i!=mark)swap(data[i],data[mark]);
return mark;
}
void quicksort(int i,int j)
{
if(i<j)
{
int q=partion(i,j);
quicksort(i,q-1);
quicksort(q+1,j);
}
}
void outputdigit(int x)
{
int out[7];
for(int i=6;i>=0;i--)
{
out[i]=x%10;
x=x/10;
}
for(int i=0;i<7;i++)
{
printf("%d",out[i]);
if(i==2)printf("-");
}
}
void output(void)
{
int num=1,i;
bool flag=false;
for(i=2;i<=data[0];i++)
{
if(data[i]==data[i-1])
{
num++;
flag=true;
}
else
{
if(num>1)
{
outputdigit(data[i-1]);
printf(" %d\n",num);
}
num=1;
}
}
if(num>1)
{
outputdigit(data[i-1]);
printf(" %d\n",num);
}
if(!flag)
printf("No duplicates.\n");
}
void input(void)
{
int n,i=1;
cin>>n;
while(n>100000)
{
cin>>n;
}
data[0]=n;
while(i<=n)
{
int num=7,temp=0;
while(num)
{
char x;
cin>>x;
if(x!='-')
{
if(x>='0'&&x<='9')
temp=temp*10+x-'0';
else if(x>='A'&&x<'Z')
{
if(x<'Q')
temp=temp*10+(x-'A')/3+2;
else temp=temp*10+(x-'Q')/3+7;
}
num--;
}
}
data[i]=temp;
i++;
}
}
int main()
{
input();
quicksort(1,data[0]);
//system("pause");
output();
//system("pause");
return 0;
}
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