本文主要是介绍CodeForces 1269C :Long Beautiful Integer,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
传送门
题目描述
分析
首先我们可以确定,答案的位数肯定是 n n n,因为所有位全部填 9 9 9是肯定符合要求的
我们把答案的前 k k k位设为 a a a的前 k k k位,设答案位 b i b_{i} bi,则 b i = b i % k b_{i}=b_{i \%k} bi=bi%k
然后判断是否合法,如果不合法的,就在 b k b_{k} bk加1即可
代码
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const int INF = 0x3f3f3f3f;
const int N = 2e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
char str[N];
int a[N],b[N];
int n,k;int main() {read(n),read(k);scanf("%s",str);for(int i = 0;i < n;i++) a[i] = str[i] - '0'; for(int i = 0;i < n;i++) b[i] = a[i];bool flag = 0;bool is = 0;for(int i = k;i < n;i++){b[i] = b[i % k];if(b[i] < a[i] && !flag){is = 1;break;}if(b[i] > a[i]) flag = 1;}if(is){b[k - 1]++;int p = k - 1;while(b[p] == 10) b[p] = 0,b[--p]++;}di(n);for(int i = 0;i < n;i++) printf("%d",b[i % k]);return 0;
}/**
* ┏┓ ┏┓+ +
* ┏┛┻━━━┛┻┓ + +
* ┃ ┃
* ┃ ━ ┃ ++ + + +
* ████━████+
* ◥██◤ ◥██◤ +
* ┃ ┻ ┃
* ┃ ┃ + +
* ┗━┓ ┏━┛
* ┃ ┃ + + + +Code is far away from
* ┃ ┃ + bug with the animal protecting
* ┃ ┗━━━┓ 神兽保佑,代码无bug
* ┃ ┣┓
* ┃ ┏┛
* ┗┓┓┏━┳┓┏┛ + + + +
* ┃┫┫ ┃┫┫
* ┗┻┛ ┗┻┛+ + + +
*/
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