The records are expressed as a string ss of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).
You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer pp is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.
In this problem, a positive integer pp is considered a period of string ss, if for all 1≤i≤|s|−p, the i-th and (i+p)-th characters of s are the same. Here |s| is the length of s.
The first line contains two space-separated integers n and p (1≤p≤n≤2000) — the length of the given string and the supposed period, respectively.
The second line contains a string ss of nn characters — Mino's records. ss only contains characters '0', '1' and '.', and contains at least one '.' character.
Output one line — if it's possible that p is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).
10 7 1.0.1.0.1.
1000100010
10 6 1.0.1.1000
1001101000
10 9 1........1
No
In the first example, 7 is not a period of the resulting string because the 1-st and 8-th characters of it are different.
In the second example, 6 is not a period of the resulting string because the 4-th and 10-th characters of it are different.
In the third example, 9 is always a period because the only constraint that the first and last characters are the same is already satisfied.
Note that there are multiple acceptable answers for the first two examples, you can print any of them.
题目:给一个串,其中 ‘ . ’ 代表未知(0/1),让你有没有办法变成一个寻环节不是p的串。
分析:仔细想想也没有什么叉点,贪心写就行啦,注意有没有访问的,随意赋值就行,别忘记这个。
代码:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n, p;
string s;
bool check(){
int len = s.length();
bool f = false;
for(int i = 0; i < len - p; i++) {
if(s[i] == '.'&&s[i+p] == '.') {
s[i] = '0';
s[i+p] = '1';
f = true;
} else if(s[i] == '.') {
if(s[i+p] == '1') s[i] = '0';
else if(s[i+p] == '0') s[i] = '1';
f = true;
} else if(s[i+p] == '.') {
if(s[i] == '1') s[i+p] = '0';
else if(s[i] == '0') s[i+p] = '1';
f = true;
} else if(s[i]!=s[i+p]) f = true;
}
if(f) for(int i = 0; i < len; i++) if(s[i] == '.') s[i] = '1';
return f;
}
int main()
{
//freopen("in.txt", "r", stdin);
while(~scanf("%d%d",&n, &p)) {
cin >> s;
bool f = false;
if(check()) f = true;
if(f) cout << s << endl;
else puts("No");
}
return 0;
}