codeforces-C. Long Beautiful Integer(字符串处理)

Codeforces Round #609 (Div. 2)-C. Long Beautiful Integer

time limit per test 3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer xx of nn digits a1,a2,…,ana1,a2,…,an, which make up its decimal notation in order from left to right.

Also, you are given a positive integer k<nk<n.

Let's call integer b1,b2,…,bmb1,b2,…,bm beautiful if bi=bi+kbi=bi+k for each ii, such that 1≤i≤m−k1≤i≤m−k.

You need to find the smallest beautiful integer yy, such that y≥xy≥x.

Input

The first line of input contains two integers n,,k (2≤n≤200000,1≤k<n2≤n≤200000,1≤k<n): the number of digits in xx and kk.

The next line of input contains nn digits a1,a2,…,ana1,a2,…,an (a1≠0a1≠0, 0≤ai≤90≤ai≤9): digits of xx.

Output

In the first line print one integer mm: the number of digits in yy.

In the next line print mm digits b1,b2,…,bmb1,b2,…,bm (b1≠0b1≠0, 0≤bi≤90≤bi≤9): digits of yy.

Examples

3 2
353

　　output

3
353

　　题意：就是给你一个长度为n的大小为十进制数x，且满足Xi=Xi+k(1≤i≤m−k.)，要你找一个大小等于y的数，其中y>=x，也要满足Yi=Yi+k(1≤i≤m−k.)。

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#include<bits/stdc++.h>
using namespace std;
char x[200005];
char y[200005];
int main(void)
{
	int a,k;
	scanf("%d %d",&a,&k);
	cin>>x;
	for(int i=0;i<a;i++){//直接模拟 
		y[i]=x[i%k];
	}
	if(strcmp(y,x)>=0)//y>=x
	{
		cout<<strlen(y)<<endl;
		cout<<y;
		return 0;
	}
	y[k-1]++;
	//如果有进位
	int i=k-1;
	while(y[i]>='9'+1)
	{
		y[i]='0';
		i--;
		y[i]++;//进位 
	}
	cout<<strlen(y)<<endl;
	for(int i=0;i<a;i++)
	y[i]=y[i%k];
	cout<<y; 
	return 0;
}