Codeforces Round #352 (Div. 2),A题与B题题解代码，水过~~

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A. Summer Camp

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Every year, hundreds of people come to summer camps, they learn new algorithms and solve hard problems.

This is your first year at summer camp, and you are asked to solve the following problem. All integers starting with 1 are written in one line.
The prefix of these line is "123456789101112131415...". Your task is to print the n-th
digit of this string (digits are numbered starting with 1.

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 1000) —
the position of the digit you need to print.

Output

Print the n-th digit of the line.

Examples

input

3

output

3

input

11

output

0

Note

In the first sample the digit at position 3 is '3',
as both integers 1 and 2 consist
on one digit.

In the second sample, the digit at position 11 is '0',
it belongs to the integer 10.

与其做BC还不如做CF，CF上的题真的是很好，虽然每次只能做两个。

这道题题意很简单，就是一连串的数字排列在一起，求第n位是几；其实一看这种题就知道是规律题，但奈何脑袋笨，只能用笨方法做，我们可以打表将这些位都存起来，数据范围不大，所以估算到400就足够有1000位了；

#include<bits/stdc++.h>
using namespace std;
const int N=1000+10;
int a[N];
int main()
{
    int n,i,j=1;
    for(i=1;i<550;i++)//其实到400就足够了；
    {
        int x=i;
        if(x<=9)
            a[j++]=x;
        else if(x<=99)//将两位数拆开存起来，下面同理；
        {
            int x1=x%10,x2=x/10;
            a[j++]=x2,a[j++]=x1;
        }
        else
        {
            int x1=x%10,x2=(x%100)/10,x3=x/100;
            a[j++]=x3,a[j++]=x2,a[j++]=x1;
        }
        if(j>=N) break;//N应该稍微开大一点，我在开始定义1000结果贡献了一个WA;
    }
    while(~scanf("%d",&n))
    {
        printf("%d\n",a[n]);
    }
    return 0;
}

其实考的就是思路，思路清晰代码能力可以这题就没什么问题了；

B. Different is Good

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A wise man told Kerem "Different is good" once, so Kerem wants all things in his life to be different.

Kerem recently got a string s consisting of lowercase English letters. Since Kerem likes it when things are different, he wants allsubstrings of
his string s to be distinct. Substring is a string formed by some number of consecutive characters of the string. For example, string
"aba" has substrings "" (empty substring), "a",
"b", "a", "ab",
"ba", "aba".

If string s has at least two equal substrings then Kerem will change characters at some positions to some other lowercase English letters.
Changing characters is a very tiring job, so Kerem want to perform as few changes as possible.

Your task is to find the minimum number of changes needed to make all the substrings of the given string distinct, or determine that it is impossible.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 100 000) —
the length of the string s.

The second line contains the string s of length n consisting
of only lowercase English letters.

Output

If it's impossible to change the string s such that all its substring are distinct print -1.
Otherwise print the minimum required number of changes.

Examples

input

2
aa

output

1

input

4
koko

output

2

input

5
murat

output

0

Note

In the first sample one of the possible solutions is to change the first character to 'b'.

In the second sample, one may change the first character to 'a' and second character to 'b',
so the string becomes "abko".

第一次看见B题这么水，哈哈，还没A题麻烦；

题意：给你一个长度为n的字符串，在其所有的子集中如果至少有两个相同的则要替换某个字母，其实单个字母就是一个子集，所以只要这个字符串中有多少个相同的就得替换多少个字母，但是，我们知道，超过26个字母无论怎么替换都有相同的，所以只要长度大于26则应该输出-1，反之求出重复字母个数输出；

至于怎么求出不同字母个数嘛；介绍一个简单函数unique(),这个我在文章分类中的知识点中已经给出了具体用法，下面来看超简洁代码：

#include<bits/stdc++.h>
using namespace std;
const int N=100000+10;
char a[N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        scanf("%s",a);
        sort(a,a+n);
        int x=unique(a,a+n)-a;
        if(n>26)
            printf("-1\n");
        else
        printf("%d\n",n-x);//输出重复的个数;
    }
    return 0;
}

目前能力只能水前两个题，请见谅；