Codeforces Global Round 1_are given an integer n n ( n ≥ 0 n≥0) represented -优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43367952/article/details/86773654

本文介绍了一种在给定基数b和位数k的情况下，通过分析数的各位数字来快速判断该数是否为偶数的方法。文章提供了一个C++实现的示例，通过读取基数b和位数k，以及各个位上的数字，然后根据基数的奇偶性和最高位数字的奇偶性来确定整个数的奇偶性。

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1 A. Parity

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
You are given an integer nn (n≥0n≥0) represented with kk digits in base (radix) bb. So,
n=a1⋅bk−1+a2⋅bk−2+…ak−1⋅b+ak.
n=a1⋅bk−1+a2⋅bk−2+…ak−1⋅b+ak.
For example, if b=17,k=3b=17,k=3 and a=[11,15,7]a=[11,15,7] then n=11⋅172+15⋅17+7=3179+255+7=3441n=11⋅172+15⋅17+7=3179+255+7=3441.
Determine whether nn is even or odd.

Input
The first line contains two integers bb and kk (2≤b≤1002≤b≤100, 1≤k≤1051≤k≤105) — the base of the number and the number of digits.
The second line contains kk integers a1,a2,…,aka1,a2,…,ak (0≤ai<b0≤ai<b) — the digits of nn.
The representation of nn contains no unnecessary leading zero. That is, a1a1 can be equal to 00 only if k=1k=1.

Output
Print “even” if nn is even, otherwise print “odd”.
You can print each letter in any case (upper or lower).

Examples
input
13 3
3 2 7
output
even
input
10 9
1 2 3 4 5 6 7 8 9
output
odd
input
99 5
32 92 85 74 4
output
odd
input
2 2
1 0
output
even

Note
In the first example, n=3⋅132+2⋅13+7=540n=3⋅132+2⋅13+7=540, which is even.
In the second example, n=123456789n=123456789 is odd.
In the third example, n=32⋅994+92⋅993+85⋅992+74⋅99+4=3164015155n=32⋅994+92⋅993+85⋅992+74⋅99+4=3164015155 is odd.
In the fourth example n=2n=2.

分析

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
const int MA=100005;
int a[MA];
int b,k;
int main()
{
    cin>>b>>k;
    int sum=0;
    for(int i=1;i<=k;++i){
        cin>>a[i];
        if(a[i]%2==1) sum++;
    }
    if(b%2==1){
      if(sum%2==1) cout<<"odd"<<endl;
      else cout<<"even"<<endl;
    }
    else {
      if(a[k]%2==1)cout<<"odd"<<endl;
      else cout<<"even"<<endl;
    }
    return 0;
}

2. B - Tape

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MA=100005;
int a[MA],b[MA];
int n,m,k,sum=0;
int main()
{
    cin>>n>>m>>k;
    for(int i=1;i<=n;++i){
        cin>>a[i];
    }
    if(n<=k){
        cout<<n<<endl;
    }
    else{
       for(int i=2;i<=n;++i){
        b[i]=a[i]-a[i-1]+1;
       }
       sort(b+2,b+n+1);
       for(int i=0;i<n-k;++i)
        sum+=b[2+i];
       cout<<sum+2*k-n<<endl;
    }
    return 0;
}