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

题目链接
题意是给定长为n的整数x，然后需要你找到不小于n并且这个数的第i位等于i+k最小的数。
很明显，可以通过把当前数x的前k位，当做循环节，查看形成的数是否大于数x，如果小于，只需把循环节加上1就能够保证形成的数大于x了，小于输出就行，这里要注意是末尾为9的时候，要有进位，当然不会出现循环节全是9加一的情况，因为全是9的形成的数一定不会小于x的。
java大数异常方便，直接加一（手动狗头）
AC代码

import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;public class Main {static StreamTokenizer st = new StreamTokenizer(new BufferedInputStream(System.in));static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));static PrintWriter pr = new PrintWriter(new BufferedOutputStream(System.out));static Scanner sc = new Scanner(System.in);public static void main(String[] args) {//      long tic = System.currentTimeMillis();int n = sc.nextInt();int k = sc.nextInt();String s = sc.next();String s2 = s.substring(0,k);boolean flag = true;for(int i = k;i<n;i++) {if(s.charAt(i)<s2.charAt((i)%k)) {break;}else if(s.charAt(i)>s2.charAt((i)%k)) {flag = false;break;}}StringBuffer sb = new StringBuffer();if(flag) {for(int i = 0;i<n;i++) {sb.append(s2.charAt(i%k));}}else{String s3 = new BigInteger(s2).add(new BigInteger("1")).toString();for(int i = 0;i<n;i++) {sb.append(s3.charAt(i%k));}}System.out.println(sb.length());System.out.println(sb);//      long toc = System.currentTimeMillis();
//      System.out.println("Elapsed time: " + (toc - tic) + " ms");}private static long A(long n, long k, long mod) {long jcn = n;for (int i = 1; i < k; i++) {jcn *= (n - i);jcn %= mod;}return jcn;}private static long C(long n, long k, long mod) {long jcn = n;long jck = 1;for (int i = 1; i < k; i++) {jcn *= (n - i);jcn %= mod;jck *= i;jck %= mod;}jck *= k;jck %= mod;return jcn * pow(jck, mod - 2, mod) % mod;}static long pow(long a, long b, long mod) {long result = 1;while (b > 0) {if (b % 2 == 1) {result = (result * a) % mod;}a = (a * a) % mod;b /= 2;}return result;}static int nextInt() {try {st.nextToken();} catch (IOException e) {e.printStackTrace();}return (int) st.nval;}static double nextDouble() {try {st.nextToken();} catch (IOException e) {e.printStackTrace();}return st.nval;}static String next() {try {st.nextToken();} catch (IOException e) {e.printStackTrace();}return st.sval;}static long nextLong() {try {st.nextToken();} catch (IOException e) {e.printStackTrace();}return (long) st.nval;}
}

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