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

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

Let’s call integer b1,b2,…,bm beautiful if bi=bi+k for each i, such that 1≤i≤m?k.

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

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

The next line of input contains n digits a1,a2,…,an (a1≠0, 0≤ai≤9): digits of x.

Output
In the first line print one integer m: the number of digits in y.

In the next line print m digits b1,b2,…,bm (b1≠0, 0≤bi≤9): digits of y.

Examples
Input
3 2
353
Output
3
353
Input
4 2
1234
Output
4
1313
思路：這個(gè)題目其實(shí)只要前k位確定了，那么這個(gè)數(shù)字就確定了。因?yàn)槿我忾L(zhǎng)度的全是9組成的數(shù)字一定符合題意，這樣m一定等于n。那么我們就取給定數(shù)字的前k位，剩下的根據(jù)這k位去構(gòu)造。如果大于等于給定的就直接輸出，否則就將前k位的數(shù)字+1，然后再根據(jù)這k位數(shù)字構(gòu)造。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;string s; int n,k;int main() {char a='a';scanf("%d%d",&n,&k);cin>>s;string t;for(int i=0;i<k;i++) t+=s[i];for(int i=k;i<n;i++) t+=t[i-k];if(t>=s) cout<<n<<endl<<t<<endl;else{for(int i=k-1;i>=0;i--){if(t[i]!='9'){t[i]+=1;break;}else t[i]='0';}for(int i=k;i<n;i++) t[i]=t[i-k];cout<<n<<endl<<t<<endl;}return 0; }

努力加油a啊，(^o)/~

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