标签：style blog color io os ar strong for 2014
A number that will be the same when it is written forwards or backwards is known as a palindromic number. For example, 1234321 is a palindromic number.
We call a number generalized palindromic number, if after merging all the consecutive same digits, the resulting number is a palindromic number. For example, 122111 is a generalized palindromic number. Because after merging, 122111 turns into 121 which is a palindromic number.
Now you are given a positive integer N, please find the largest generalized palindromic number less than N.
There are multiple test cases. The first line of input contains an integer T (about 5000) indicating the number of test cases. For each test case:
There is only one integer N (1 <= N <= 1018).
For each test case, output the largest generalized palindromic number less than N.
4 12 123 1224 1122
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 401
#define MAXN 200005
#define INF 0x3f3f3f3f
#define mod 1000000007
#define eps 1e-6
const double pi=acos(-1.0);
typedef long long ll;
using namespace std;
ll n,m,ans;
int tot,dig[20],le[20],ri[20];
ll get(int num1,int num2)
{
int i;
ll res=0;
for(i=1;i<=num1;i++)
{
res=res*10+le[i];
}
for(i=num2;i>=1;i--)
{
res=res*10+ri[i];
}
return res;
}
ll dfs(int num1,int num2,int limit)
{
ll t,res=0;
if(num1+num2>tot)
{
res=get(num1-1,num2);
if(res<m) return res;
return 0;
}
int i,j,ed;
ed=limit?dig[tot-num1+1]:9;
for(i=ed;i>=0;i--)
{
le[num1]=i;
if((num1==1||le[num1]!=le[num1-1])&&!(num1==1&&i==0)&&(num1+num2<tot))
{
for(j=1; num1+num2+j<=tot; j++) // r与之匹配的个数
{
ri[num2+j]=i;
t=dfs(num1+1,num2+j,limit&&(i==ed));
res=max(res,t);
}
}
else
{
t=dfs(num1+1,num2,limit&&(i==ed));
res=max(res,t);
}
if(res>0) return res;
}
return res;
}
void solve()
{
int i,j;
tot=0;
m=n;
while(n)
{
dig[++tot]=n%10;
n/=10;
}
ans=dfs(1,0,1);
printf("%lld\n",ans);
}
int main()
{
int i,j,t;
scanf("%d",&t);
while(t--)
{
scanf("%lld",&n);
solve();
}
return 0;
}zoj 3816 Generalized Palindromic Number （根据对称性来搜）
标签：style blog color io os ar strong for 2014
原文地址：http://blog.csdn.net/tobewhatyouwanttobe/article/details/39227223
