CF1268A Long Beautiful Integer(贪心)

传送门

思路:

贪心的考虑,如果将原数组的前$k$位放满$n$位能够满足条件的话,就会这样放。

如果不能满足条件的话,将该$k$位$+1$后一定会满足条件,这便是最小的。

要注意进位的情况,并且如果是9999这种情况的话,要变成10000,。

代码:

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=5e5+100;

int n,k,a[210000],b[210000];

bool check(){
    rep(i,1,n){
        if(a[i]<b[i]) return 1;
        else if(a[i]>b[i]) return 0;
    }
    return 1;
}

int main()
{
    n=read,k=read;
    rep(i,1,n) scanf("%1d",&a[i]);
    rep(i,1,n){
        if(i<=k) b[i]=a[i];
        else b[i]=b[i-k];
    }
    if(check()){
        cout<<n<<endl;
        rep(i,1,n) cout<<b[i];
        puts("");
    }
    else{
        b[k]++;
        for(int i=k;i;i--){
            if(b[i]>9){
                b[i]%=10;b[i-1]++;
            }
            else break;
        }
        if(b[0]==1){
            k++;n+=2;
            b[1]=1;
            rep(i,2,k) b[i]=0;
        }
        rep(i,1+k,n){
            b[i]=b[i-k];
        }
        cout<<n<<endl;
        rep(i,1,n) cout<<b[i];
    }
    return 0;
}

/*

**/

 

posted @ 2021-06-06 14:07  OvO1  阅读(69)  评论(0编辑  收藏  举报