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.
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.
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.)。
#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;
}
codeforces-C. Long Beautiful Integer(字符串处理)
原文：https://www.cnblogs.com/xuanmaiboy/p/12105242.html