CodeForces 260A Adding Digits

这道题目的意思是给你提供a, b, n 三个数

a为 输入的数字 ，你需要在ａ后面加ｎ次 ，每次可以加0-9

但要保证每次加上去的那个数字能被b整除  

不过数据规模有点大，用搜索会MLE(即使开了个开栈挂

#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int N = 1              ;
const int M = 200000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;

string a;
int b, n;
bool flag = false;

string Multiply(string s,int x)  //大数乘以整形数
{
    reverse(s.begin(),s.end());
    int cmp=0;
    for(int i=0;i<s.size();i++)
    {
        cmp=(s[i]-'0')*x+cmp;
        s[i]=(cmp%10+'0');
        cmp/=10;
    }
    while(cmp)
    {
        s+=(cmp%10+'0');
        cmp/=10;
    }
    reverse(s.begin(),s.end());
    return s;
}

string Except(string s,int x)  //大数除以整形数
{
    int cmp=0,ok=0;
    string ans="";
    for(int i=0;i<s.size();i++)
    {
        cmp=(cmp*10+s[i]-'0');
        if(cmp>=x)
        {
            ok=1;
            ans+=(cmp/x+'0');
            cmp%=x;
        }
        else{
            if(ok==1)
                ans+='0';  //注意这里啊。才找出错误
        }
    }
    return ans;
}

void dfs(string a, int len){
    if(flag)    return;
    if(n == len){
        cout << a << endl;
        flag = true;
        return;
    }
    for(int i = 0; i < 10; ++i){
        string num = a;
        num.push_back('0' + i);
        string nn = Except(num, b);
        string aa = Multiply(nn, b);
        if(aa.compare(num) == 0){
            dfs(num, len + 1);
        }
    }
}

int main(){
    int i, j, k, t, m, numCase = 0;
    //string hh = "123";
    //cout << atoi(hh.c_str());
    while(cin >> a >> b >> n){
        dfs(a, 0);
        if(!flag){
            cout << "-1" << endl;
        }

    }

    return 0;
}

解题思路：

非常巧妙，就是先在a后面加一个数字0-9如果能整除b，那么易得an后面跟无论多少个0都能整除b

所以就在an后面构造出(n-1)个0即可

无须搜索，超过30层容易爆空间~

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int N = 1              ;
const int M = 200000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;



int main(){
    int i, j, k, t, n, m, numCase = 0;
    int a, b, num;
    //string hh = "123";
    //cout << atoi(hh.c_str());
    while(cin >> a >> b >> n){
            bool flag = false;
            for(i = 0; i < 10; ++i){
                num = a * 10 + i;
                if(num % b == 0){
                    a = num;
                    flag = true;
                    break;
                }
            }
            if(!flag){
                cout << -1 << endl;
                return 0;
            }
        cout << num;
        for(i = 1; i < n; ++i) cout << '0';
        cout << endl;
    }

    return 0;
}