微软2016校园招聘在线笔试第二场

题目2 : Numeric Keypad

时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

The numberic keypad on your mobile phone looks like below:

1 2 3
4 5 6
7 8 9
0

Suppose you are holding your mobile phone with single hand. Your thumb points at digit 1. Each time you can 1) press the digit your thumb pointing at, 2) move your thumb right, 3) move your thumb down. Moving your
thumb left or up is not allowed.
By using the numeric keypad under above constrains, you can produce some numbers like 177 or 480 while producing other numbers like 590 or 52 is impossible.
Given a number K, find out the maximum number less than or equal to K that can be produced.

输入

The first line contains an integer T, the number of testcases.
Each testcase occupies a single line with an integer K.

For 50% of the data, 1 <= K <= 999.
For 100% of the data, 1 <= K <= 10500, t <= 20.

输出

For each testcase output one line, the maximum number less than or equal to the corresponding K that can be produced.

样例输入

3
25
83
131

样例输出

25
80
129

#include<iostream>
#include<string>
#include<vector>
using namespace std;

int inside(int k, vector<int> nextstep)
{
int flag;
int len = nextstep.size();
int i;
if (k < nextstep[0])
return -1;
for (i = 0; i < len; i++)
{

if (k<nextstep[i])
break;
else
flag = nextstep[i];
}

return flag;
}

int maxiNum(vector<vector<int>> next,string K)
{
int res=0;
int len = K.length();
int flag = 0;
int i,j;
int temp;
j=1;
for (i = 0; i < len; i++)
{
temp = inside(K[i] - '0', next[j]);

if (flag == 1)
{
res *= 10;
j = next[j][next[j].size() - 1];
res += j;
}
else if (temp == -1)
{
do{
temp = res % 10-1;
res /= 10;

if ((inside(temp, next[res%10]))!=-1||res==0)
break;
else
i--;
} while (1);
res *= 10;
res += temp;
flag = 1;
i--;
}
else
{
res *= 10;
j = temp;
res += j;
if (temp != K[i] - '0')
{
flag = 1;
}
}

}

return res;
}

int main()
{
int T;
string K;
vector<int> res;
int i;

vector<vector<int>> next;
vector<int> nextstep;

nextstep.push_back(0);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(0);
nextstep.push_back(1);
nextstep.push_back(2);
nextstep.push_back(3);
nextstep.push_back(4);
nextstep.push_back(5);
nextstep.push_back(6);
nextstep.push_back(7);
nextstep.push_back(8);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(0);
nextstep.push_back(2);
nextstep.push_back(3);
nextstep.push_back(5);
nextstep.push_back(6);
nextstep.push_back(8);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(3);
nextstep.push_back(6);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(0);
nextstep.push_back(4);
nextstep.push_back(5);
nextstep.push_back(6);
nextstep.push_back(7);
nextstep.push_back(8);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(0);
nextstep.push_back(5);
nextstep.push_back(6);
nextstep.push_back(8);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(6);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(0);
nextstep.push_back(7);
nextstep.push_back(8);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(0);
nextstep.push_back(8);
nextstep.push_back(9);
next.push_back(nextstep);

nextstep.clear();
nextstep.push_back(9);
next.push_back(nextstep);

cin >> T;
for (i = 0; i < T; i++)
{
cin >> K;
res.push_back(maxiNum(next,K));
}
for (i = 0; i < T; i++)
cout << res[i] << endl;

return 0;
}