CF989B A Tide of Riverscape 思维 第七题

A Tide of Riverscape

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Walking along a riverside, Mino silently takes a note of something.

"Time," Mino thinks aloud.

"What?"

"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."

"And what are you recording?"

"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.

Doubtfully, Kanno peeks at Mino's records.

The records are expressed as a string $\mathrm{ss of\; characters\; \text{'}0\text{'},\; \text{'}1\text{'}\; and\; \text{'}.\text{'},\; where\; \text{'}0\text{'}\; denotes\; a\; low\; tide,\; \text{'}1\text{'}\; denotes\; a\; high\; tide,\; and\; \text{'}.\text{'}\; denotes\; an\; unknown\; one\; (either\; high\; or\; low).}$

You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $\mathrm{pp is not a\; period\; of\; the\; resulting\; string.\; In\; case\; the\; answer\; is\; yes,\; please\; also\; show\; such\; a\; replacement\; to\; Mino.}$

In this problem, a positive integer $\mathrm{pp is\; considered\; a\; period\; of\; string ss,\; if\; for\; all 1\le i\le |s|-p1\le i\le |s|-p,\; the ii-th\; and (i+p)(i+p)-th\; characters\; of ssare\; the\; same.\; Here |s||s| is\; the\; length\; of ss.}$

Input

The first line contains two space-separated integers $\mathrm{nn and pp (1\le p\le n\le 20001\le p\le n\le 2000) —\; the\; length\; of\; the\; given\; string\; and\; the\; supposed\; period,\; respectively.}$

The second line contains a string $\mathrm{ss of nn characters —\; Mino\text{'}s\; records. ss only\; contains\; characters\; \text{'}0\text{'},\; \text{'}1\text{'}\; and\; \text{'}.\text{'},\; and\; contains\; at\; least\; one\; \text{'}.\text{'}\; character.}$

Output

Output one line — if it's possible that $\mathrm{pp is not a\; period\; of\; the\; resulting\; string,\; output\; any\; one\; of\; such\; strings;\; otherwise\; output\; "No"\; (without\; quotes,\; you\; can\; print\; letters\; in\; any\; case\; (upper\; or\; lower)).}$

Examples

input

Copy

10 7
1.0.1.0.1.

output

Copy

1000100010

input

Copy

10 6
1.0.1.1000

output

Copy

1001101000

input

Copy

10 9
1........1

output

Copy

No

Note

In the first example, $\mathrm{77 is\; not\; a\; period\; of\; the\; resulting\; string\; because\; the 11-st\; and 88-th\; characters\; of\; it\; are\; different.}$

In the second example, $\mathrm{66 is\; not\; a\; period\; of\; the\; resulting\; string\; because\; the 44-th\; and 1010-th\; characters\; of\; it\; are\; different.}$

In the third example, $\mathrm{99 is\; always\; a\; period\; because\; the\; only\; constraint\; that\; the\; first\; and\; last\; characters\; are\; the\; same\; is\; already\; satisfied.}$

Note that there are multiple acceptable answers for the first two examples, you can print any of them.

 啊啊啊啊啊啊，简直了，佩服我自己，一个等于符号写成赋值符号，调半天bug

题意：给你n个字符串，问你每隔k个是否右不相等的，没有输出"No"，有输出一个符合不相等的（注意‘.'可以是0或者1，所以输出的时候'.'是会根据前后的数字发生变化的）

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int main(){
    std::ios::sync_with_stdio(false);
    ll n, k;
    while( cin >> n >> k ) {
        string s;
        cin >> s;
        ll num;
        bool flag = false;
        for( ll i = 0; i < s.length(); i ++ ) {
            if( i + k < s.length() ) {
                if( s[i] != s[i+k] || ( s[i] == s[i+k] && s[i] == '.' ) ) {
                    if( s[i] == s[i+k] ) {
                        s[i] = '0', s[i+k] = '1';
                    } else {
                        if( s[i] == '0' ) {
                            s[i+k] = '1';
                        } else if( s[i] == '1' ) {
                            s[i+k] = '0';
                        } else {
                            if( s[i+k] == '0' ) {  //这里等于号写成了赋值符号，调了半天bug，吐血ing
                                s[i] = '1';
                            } else {
                                s[i] = '0';
                            }
                        }
                    }
                    flag = true;
                    break;
                }
            }
        }
        if( flag ) {
            for( ll i = 0; i < s.length(); i ++ ) {
                if( s[i] == '.' ) {
                    cout << "0";
                } else {
                    cout << s[i];
                }
            }
        } else {
            cout << "No";
        }
        cout << endl;
    }
    return 0;
}