A. Odds and Ends
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?

Given an integer sequence a1, a2, ..., an of length n. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.

A subsegment is a contiguous slice of the whole sequence. For example, {3, 4, 5} and {1} are subsegments of sequence {1, 2, 3, 4, 5, 6}, while {1, 2, 4} and {7} are not.

Input

The first line of input contains a non-negative integer n (1 ≤ n ≤ 100) — the length of the sequence.

The second line contains n space-separated non-negative integers a1, a2, ..., an (0 ≤ ai ≤ 100) — the elements of the sequence.

Output

Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.

You can output each letter in any case (upper or lower).

Examples
input 
31 3 5

output 
Yes

input 
51 0 1 5 1

output 
Yes

input 
34 3 1

output 
No

input 
43 9 9 3

output 
No

Note

In the first example, divide the sequence into 1 subsegment: {1, 3, 5} and the requirements will be met.

In the second example, divide the sequence into 3 subsegments: {1, 0, 1}, {5}, {1}.

In the third example, one of the subsegments must start with 4 which is an even number, thus the requirements cannot be met.

In the fourth example, the sequence can be divided into 2 subsegments: {3, 9, 9}, {3}, but this is not a valid solution because 2 is an even number.

问你把长度为n的序列化为奇数开头奇数结尾而且个数是奇数的数列

首先分析偶数，凑两个奇数数列，然后这个子序列个数就又是偶数了

所以只有奇数个符合要求，开头结尾必须是，合并为一个就好了

我错了是因为优先级 判断末位是不是奇数要用 (n&1)==0

#include<bits/stdc++.h>
using namespace std;
int main()
{int n;cin>>n;int a[102];for(int i=1;i<=n;i++)cin>>a[i];int f=0;if(n&1&&a[1]&1&&a[n]&1)f=1;printf("%s",f?"Yes":"No");return 0;
}

B. Tell Your World
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Connect the countless points with lines, till we reach the faraway yonder.

There are n points on a coordinate plane, the i-th of which being (i, yi).

Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on exactly one of them, and each of them passes through at least one point in the set.

Input

The first line of input contains a positive integer n (3 ≤ n ≤ 1 000) — the number of points.

The second line contains n space-separated integers y1, y2, ..., yn ( - 109 ≤ yi ≤ 109) — the vertical coordinates of each point.

Output

Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.

You can print each letter in any case (upper or lower).

Examples
input 
57 5 8 6 9

output 
Yes

input 
5-1 -2 0 0 -5

output 
No

input 
55 4 3 2 1

output 
No

input 
51000000000 0 0 0 0

output 
Yes

Note

In the first example, there are five points: (1, 7), (2, 5), (3, 8), (4, 6) and (5, 9). It's possible to draw a line that passes through points 1, 3, 5, and another one that passes through points 2, 4 and is parallel to the first one.

In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.

In the third example, it's impossible to satisfy both requirements at the same time.

这个题嘛就是给你一堆点，让你判断在不在两条平行（不重合）的直线上

我选用前三个点在不在一条直线上，在的话就是其他点还有一条直线

不在的话就分三种情况讨论

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a[1005];int n;
bool check()
{int f=-1;LL f1=a[3]-a[2];if(a[2]*2==a[1]+a[3]){for(int i=4; i<=n; i++){if((a[i]-a[1])!=(i-1)*f1){if(f==-1)f=i;else{if((a[i]-a[f])!=(i-f)*f1){return 0;}}}}if(f==-1){return 0;}}else{f=0,f1=a[2]-a[1];for(int i=4; i<=n; i++){if((a[i]-a[1])!=(i-1)*f1&&((a[i]-a[3])!=(i-3)*f1)){f++;break;}}f1=a[3]-a[2];for(int i=4; i<=n; i++){if((a[i]-a[2])!=(i-2)*f1&&((a[i]-a[1])!=(i-1)*f1)){f++;break;}}f1=a[3]-a[1];for(int i=4; i<=n; i++){if(2*(a[i]-a[2])!=(i-2)*f1&&(2*(a[i]-a[1])!=(i-1)*f1)){f++;break;}}if(f==3)return 0;}return 1;
}
int main()
{scanf("%d",&n);for(int i=1; i<=n; i++)scanf("%lld",a+i);if(!check())printf("No\n");else printf("Yes\n");return 0;
}

这个写法好啊 
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1005;
ll y[N];
int n;
bool gx(int a,int b,int c,int d)
{return (b-a)*(y[d]-y[c])==(d-c)*(y[b]-y[a]);
}
int check(int d,int f)
{vector<int>V;for(int i=1; i<=n; i++)if(!gx(d,f,f,i))V.push_back(i);if(V.size()==0)return 0;if(V.size()==1)return 1;for(int i=1; i<(int)V.size(); i++)if(!gx(d,f,V[i],V[i-1]))return 0;return 1;
}
int main()
{cin>>n;for(int i=1; i<=n; i++)cin>>y[i];if(check(1,2)||check(2,3)||check(1,3))cout<<"YES";else cout<<"NO";return 0;
}

C. From Y to Y
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

From beginning till end, this message has been waiting to be conveyed.

For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length 1, and repeat the following operation n - 1 times:

  • Remove any two elements s and t from the set, and add their concatenation s + t to the set.

The cost of such operation is defined to be , where f(s, c) denotes the number of times character c appears in string s.

Given a non-negative integer k, construct any valid non-empty set of no more than 100 000 letters, such that the minimum accumulative cost of the whole process is exactly k. It can be shown that a solution always exists.

Input

The first and only line of input contains a non-negative integer k (0 ≤ k ≤ 100 000) — the required minimum cost.

Output

Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.

Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.

Examples
input 
12

output 
abababab

input 
3

output 
codeforces

Note

For the multiset {'a', 'b', 'a', 'b', 'a', 'b', 'a', 'b'}, one of the ways to complete the process is as follows:

  • {"ab", "a", "b", "a", "b", "a", "b"}, with a cost of 0;
  • {"aba", "b", "a", "b", "a", "b"}, with a cost of 1;
  • {"abab", "a", "b", "a", "b"}, with a cost of 1;
  • {"abab", "ab", "a", "b"}, with a cost of 0;
  • {"abab", "aba", "b"}, with a cost of 1;
  • {"abab", "abab"}, with a cost of 1;
  • {"abababab"}, with a cost of 8.

The total cost is 12, and it can be proved to be the minimum cost of the process.

就是贪心构造，合并花费是0，但是连续的某个字母生成的话费是n*（n-1）/2，每次生成最多就好的

#include <bits/stdc++.h>
using namespace std;
int a[500];
int main()
{int k;scanf("%d",&k);for(int i=1;i<450;i++)a[i]=i*(i-1)/2;if(k==0)printf("a");else{int f=0;char c='a';while(k){int pos=lower_bound(a+1,a+450,k)-a;if(a[pos]>k)pos--;for(int i=0;i<pos;i++)printf("%c",c);k-=a[pos];c++;}}return 0;
}

转载于:https://www.cnblogs.com/BobHuang/p/7465836.html

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