A. Odds and Ends

简单判断

#include<cstdio>
int a[105];

int main(){
	int n;
	while(~scanf("%d",&n)){
		int cnt=0;
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		
		
		if(n%2==0||a[0]%2==0||a[n-1]%2==0)
			printf("No\n");
				
		if(n%2&&a[0]%2&&a[n-1]%2)
			printf("Yes\n");
			
	}
	return 0;
} 

B. Tell Your World

斜率k是点1，2，3构成三种直线的其一斜率，然后寻找两条等斜率的基准点逐点判断

#include<cstdio>
const int N=1005;
int p[N];
int n;

bool work(double k){
	int flag=0,next=-1;
	for(int i=2;i<=n;i++){
		if(p[i]-p[1]==k*(i-1)) continue; //1,i
		flag=1;  //other
		if(next<0) next=i;  
		else if(p[i]-p[next]!=k*(i-next)){
			flag=0;
			break;
		}
	}
	if(flag) return true;
	else return false;
}

int main(){
	while(~scanf("%d",&n)){
		for(int i=1;i<=n;i++)
			scanf("%d",&p[i]);
		
		double k1=1.0*(p[2]-p[1]);  //1,2
		double k2=0.5*(p[3]-p[1]);  //1,3
		double k3=1.0*(p[3]-p[2]);  //2,3
		
		if(work(k1)||work(k2)||work(k3))
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}

C. From Y to Y

一个字符串中有n个a的话，那么对应这些字符a贡献的价值是：（n*(n-1)）/2

#include<cstdio>
#include<cstring>
const int N=100;

int main(){
	int n;
	while(~scanf("%d",&n)){
		int cost=0;
		for(int i=1;i<=26;i++){
			int pos=100;
			for(int j=1;j<=N;j++){
				if(cost+j*(j-1)/2>n){
					pos=j-1;
					break;
				}
			}
			
			for(int k=1;k<=pos;k++)
				printf("%c",i+'a'-1);
			cost+=(pos-1)*pos/2;
			if(cost==n) break;
		}
		printf("\n");
	}
	return 0;
}