Problem Statement
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Given two positive integers x and y, their similarity S(x, y) is defined as follows: To compute S(x, y) we count all d between 0 and 9, inclusive, such that both x and y contain the digit d when written in base 10 (without any leading zeros). For example,
S(1123, 220181) = 2 since both numbers contain the digit 1 and both contain the digit 2.
You are given two ints L and R that define a range. Find two distinct integers in this range that have the largest similarity. Formally, return the maximum of S(a, b) over all a, b such thatL <= a < b <=
R.
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Definition
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| Class: |
Similars |
| Method: |
maxsim |
| Parameters: |
int, int |
| Returns: |
int |
| Method signature: |
int maxsim(int L, int R) |
| (be sure your method is public) |
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Limits
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| Time limit (s): |
2.000 |
| Memory limit (MB): |
256 |
| Stack limit (MB): |
256 |
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Constraints
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R will be between 2 and 100,000, inclusive.
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L will be between 1 and R - 1, inclusive.
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Examples
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| 0) |
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Returns: 1
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| We have S(1, 10) = 1 since both numbers contain the digit 1. All other pairs of numbers within this range have similarity 0. |
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| 1) |
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Returns: 2
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| There are many pairs with similarity 2, for example pairs (23,32) and (38,83). |
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| 2) |
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Returns: 0
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| Here we have only one pair (99, 100) and its similarity is 0. |
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| 3) |
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| 4) |
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直接把每个数根据数位状压成一个数，共2^10 统计即可
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MEM(a) memset(a,0,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000)
typedef long long ll;
int f[MAXN]={0},g[MAXN]={0};
int p2[20]={1,2,4,8,16,32,64,128,256,512,1024};
class Similars
{
public:
int maxsim(int L, int R)
{
int ans=0;
p2[0]=1;
For(i,10) p2[i]=p2[i-1]*2;
MEM(f)
Fork(i,L,R)
{
int p=i,t=0;
while (p)
{
t=t|p2[p%10];
p/=10;
}
f[t]++;
}
Rep(i,p2[10])
{
int q=0;
Rep(j,10) if (i&p2[j]) q++;
g[i]=q;
}
Rep(i,p2[10])
Rep(j,p2[10])
{
if ( (i==j && f[i]>=2) || (i!=j && f[i] && f[j] ))
{
ans=max(ans,g[i&j]);
}
}
return ans;
}
};
时间： 2024-12-17 20:43:32
