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Re:这题也可以爆搜的In Reply To:Re:这题也可以爆搜的 Posted by:vjudge110 at 2016-03-05 23:00:52 你们爆搜水过 我10!的复杂度爆搜竟然超时。。。。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
int minn=999999999;
int m;
char a[11]={0};
char perm1[11]={0},perm2[11]={0};
int powa(int i,int j)
{
int powar=i;
if(j>0)
{
for(int k=1;k<=j-1;k++)
{
powar*=i;
}
return powar;
}
else if(j==0)
{
powar=1;
return powar;
}
}
int sum(char perm[])
{
int len=strlen(perm);
int num=0;
int jlen=len;
while(jlen>=1)
{
for(int i=0;i<len;i++)
{
int t=jlen-1;
num+=(perm[i]-'0')*powa(10,t);
jlen--;
}
}
return num;
}
int main()
{
//freopen("1.in","r",stdin);
//freopen("1.out","w",stdout);
scanf("%d\n",&m);
char ch;
for(int i=0;i<m;i++)
{
minn=99999999;
int k;
int tag=0;
while((ch=getchar())!='\n')
{
if(ch!=' ')
{
a[tag]=ch;
tag++;
}
}
int len=strlen(a);
do
{
int flag=0;
if(len%2==1)
{
for(k=0;k<len/2+1;k++)
{
perm1[k]=a[k];
}
for(int j=k;j<len;j++)
{
perm2[flag]=a[j];
flag++;
}
if(perm1[0]!='0' && perm2[0]!='0')
{
int tag1=sum(perm1);
int tag2=sum(perm2);
int del=abs(tag1-tag2);
if(minn>del)
{
minn=del;
}
}
}
else if(len%2==0)
{
for(k=0;k<len/2;k++)
{
perm1[k]=a[k];
}
for(int j=k;j<len;j++)
{
perm2[flag]=a[j];
flag++;
}
if(perm1[0]!='0' && perm2[0]!='0' && perm2[0]-perm1[0]==1)
{
int tag1=sum(perm1);
int tag2=sum(perm2);
int del=abs(tag1-tag2);
if(minn>del)
{
minn=del;
}
}
}
}while(next_permutation(a,a+len));
printf("%d\n",minn);
}
return 0;
}
正在优化。。
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