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A. Giga Tower
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888 . In particular, there is floor 0 between floor - 1 and floor 1 . Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky , if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a . He wants to find the minimum positive integer b , such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 10 9 ≤ a ≤ 10 9 ).
Output
Print the minimum b in a line.
Sample test(s)
input
179
output
1
input
-1
output
9
input
18
output
10
#include <iostream>
#include <algorithm>
#include <iterator>
#include <deque>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#include <set>
#include <valarray>
#include <list>
#include <stack>
#include <array>
#include <iomanip>
#include <map>
#include <string>
#include <queue>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <stdio.h>
using namespace std;
long long n,m,x;
int check (long long a)
{
if( a < 0 )
a = -a;
while ( a )
{
if (a % 10 == 8 )
return 1 ;
a /= 10;
}
return 0;
}
int main()
{
while (scanf("%I64d",&n)!=EOF)
{
int ans =0 ;
while (true)
{
ans ++;
n++;
if (check (n))
{
printf("%d\n",ans);
break;
}
}
}
return 0;
}
B暴力
http://codeforces.com/contest/488/problem/B
B. Candy Boxes
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
There is an old tradition of keeping 4 boxes of candies in the house in
Cyberland. The numbers of candies are special if their arithmetic mean ,
their median and their range are all equal. By definition, for a set {
x 1 , x 2 , x 3 , x 4 } ( x 1 ≤ x 2 ≤ x 3 ≤ x 4 )
arithmetic mean is
, median is
and range is x 4 - x 1 . The arithmetic mean and median are not
necessary integer. It is well-known that if those three numbers are same,
boxes will create a "debugging field" and codes in the field will have no
bugs.
For example, 1, 1, 3, 3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2 .
Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only n ( 0 ≤ n ≤ 4 ) boxes remaining. The i -th remaining box contains a i candies.
Now Jeff wants to know: is there a possible way to find the number of candies of the 4 - n missing boxes, meeting the condition above (the mean, median and range are equal)?
Input
The first line of input contains an only integer n ( 0 ≤ n ≤ 4 ).
The next n lines contain integers a i , denoting the number of candies in the i -th box ( 1 ≤ a i ≤ 500 ).
Output
In the first output line, print " YES " if a solution exists, or print " NO " if there is no solution.
If a solution exists, you should output 4 - n more lines, each line containing an integer b , denoting the number of candies in a missing box.
All your numbers b must satisfy inequality 1 ≤ b ≤ 10 6 . It is guaranteed that if there exists a positive integer solution, you can always find such b 's meeting the condition. If there are multiple answers, you are allowed to print any of them.
Given numbers a i may follow in any order in the input, not necessary in non-decreasing.
a i may have stood at any positions in the original set, not necessary on lowest n first positions .
Sample test(s)
input
2
1
1
output
YES
3
3
input
3
1
1
1
output
NO
input
4
1
2
2
3
output
YES
Note
For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3 . The arithmetic mean, the median and the range of them are all 2 .
For the second sample, it's impossible to find the missing number of candies.
In the third example no box has been lost and numbers satisfy the condition.
You may output b in any order.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<set>
#include<utility>
#define inf 0x7fffffff
using namespace std;
const int maxn=100+10;
int main()
{
int n;
int an[5];
while (scanf("%d",&n)!=EOF)
{
for (int i=1 ;i<=n ;i++) scanf("%d",&an[i]);
sort(an+1,an+n+1);
if (n==0)
{
printf("YES\n");
printf("1\n1\n3\n3\n");continue;
}
if (n==1)
{
int t=2*an[1];
printf("YES\n");
printf("%d\n%d\n%d\n",t/2,t/2*3,t/2*3);
continue;
}
if (n==2)
{
int flag=0;
// 1 2
int t=2*an[1];
int bn[5];
bn[1]=an[1],bn[4]=3*bn[1];
bn[2]=an[2];
bn[3]=4*bn[1]-bn[2];
if (bn[2]>=bn[1] && bn[3]>=bn[2] && bn[4]>=bn[3] && bn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%d\n%d\n",bn[3],bn[4]);
continue;
}
// 1 3
t=2*an[1];
bn[1]=an[1];
bn[3]=an[2];
bn[2]=2*t-bn[3];
bn[4]=bn[1]+t;
if (bn[2]>=bn[1] && bn[3]>=bn[2] && bn[4]>=bn[3] && bn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%d\n%d\n",bn[2],bn[4]);
continue;
}
// 1 4
t=2*an[1];
bn[1]=an[1];
bn[4]=bn[1]+t;
if (an[2]==bn[4])
{
bn[2]=bn[1];
bn[3]=bn[4];
if (bn[2]>=bn[1] && bn[3]>=bn[2] && bn[4]>=bn[3] && bn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%d\n%d\n",bn[2],bn[3]);
continue;
}
}
// 2 3
double cn[5];
double tt=(double)(an[1]+an[2])/2;
cn[2]=an[1] ;cn[3]=an[2] ;
cn[1]=tt/2.0;
cn[4]=1.5*tt;
if (cn[2]>=cn[1] && cn[3]>=cn[2] && cn[4]>=cn[3] && cn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%.0lf\n%.0lf\n",cn[3],cn[4]);
continue;
}
// 2 4
cn[4]=an[2];
cn[2]=an[1];
tt=2.0*cn[4]/3.0;
cn[1]=tt/2.0;
cn[3]=2.0*tt-cn[2];
if (cn[2]>=cn[1] && cn[3]>=cn[2] && cn[4]>=cn[3] && cn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%.0lf\n%.0lf\n",cn[1],cn[3]);
continue;
}
// 3 4
cn[3]=an[1];
cn[4]=an[2];
tt=2.0*cn[4]/3.0;
cn[1]=tt/2.0;
cn[2]=2.0*tt-cn[3];
if (cn[2]>=cn[1] && cn[3]>=cn[2] && cn[4]>=cn[3] && cn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%.0lf\n%.0lf\n",cn[1],cn[2]);
continue;
}
if (!flag) printf("NO\n");
}
if (n==3)
{
// 1 2 3
double cn[5];
int flag=0;
cn[1]=an[1] ;cn[2]=an[2] ;cn[3]=an[3] ;
double tt=2.0*cn[1];
cn[4]=1.5*tt;
if (cn[2]+cn[3]==2.0*tt)
{
if (cn[2]>=cn[1] && cn[3]>=cn[2] && cn[4]>=cn[3] && cn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%.0lf\n",cn[4]);
continue;
}
}
// 1 2 4
cn[1]=an[1] ;cn[2]=an[2] ;cn[4]=an[3];
tt=2.0*cn[1];
cn[3]=2.0*tt-cn[2];
if (2.0*cn[4]==3.0*tt)
{
if (cn[2]>=cn[1] && cn[3]>=cn[2] && cn[4]>=cn[3] && cn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%.0lf\n",cn[3]);
continue;
}
}
// 2 3 4
cn[2]=an[1] ;cn[3]=an[2] ;cn[4]=an[3] ;
tt=(cn[2]+cn[3])/2.0;
cn[1]=tt/2.0;
if (cn[4]*2.0==3.0*tt)
{
if (cn[2]>=cn[1] && cn[3]>=cn[2] && cn[4]>=cn[3] && cn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%.0lf\n",cn[1]);
continue;
}
}
// 1 3 4
cn[1]=an[1] ;cn[3]=an[2] ;cn[4]=an[3] ;
tt=2.0*cn[1];
cn[2]=2.0*tt-cn[3];
if (2.0*cn[4]==3.0*tt)
{
if (cn[2]>=cn[1] && cn[3]>=cn[2] && cn[4]>=cn[3] && cn[4]<=1000000)
{
flag=1;
printf("YES\n");
printf("%.0lf\n",cn[2]);
continue;
}
}
if (!flag) printf("NO\n");
}
if (n==4)
{
double tt=2.0*an[1];
int flag=0;
if (an[2]+an[3]==2.0*tt && 2.0*an[4]==3.0*tt)
{
if (an[2]>=an[1] && an[3]>=an[2] && an[4]>=an[3] && an[4]<=1000000)
{
flag=1;
printf("YES\n");
continue;
}
}
if (!flag) printf("NO\n");
}
}
return 0;
}
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