Codeforces 989A:A Blend of Springtime
1 second
256 megabytes
standard input
standard output
"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."
"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.
The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.
When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.
You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.
The first and only line of input contains a non-empty string ss consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only (|s|≤100|s|≤100) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.
Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.
You can print each letter in any case (upper or lower).
- .BAC.
- Yes
- AA..CB
- No
In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.
In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.
题意:找一个字符串中是否有连续的三个字符,这三个字符都不相等,且不为"."//
//WA了好多发都是因为大意,没仔细读写好的代码,难受
#include <bits/stdc++.h>
using namespace std;
const int maxn=1000;
char ch[maxn];
int main(int argc, char const *argv[])
{
cin>>ch;
int l=strlen(ch);
int flag=0;
for(int i=0;i<l-2;i++)
{
if(ch[i]!=ch[i+1]&&ch[i+1]!=ch[i+2]&&ch[i+2]!=ch[i]&&ch[i]!='.'&&ch[i+1]!='.'&&ch[i+2]!='.')
flag++;
}
if(flag)
cout<<"Yes"<<endl;
else
cout<<"No"<<endl;
return 0;
}
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