E Easy problem

链接：https://ac.nowcoder.com/acm/contest/338/E
来源：牛客网

Zghh likes number, but he doesn't like writing problem description. So he will just give you a problem instead of telling a long story for it.
Now given a positive integer x and k digits a1,a2,...,ak, can you find a positive integer y such that y is the multiple of x and in decimal representation y contains all digits of a1,a2,...,ak.

输入描述:

The first line contains an integer T (1<=T<=10000) which is the number of test case.The following T lines each line is a test case, start with two integer x (1<=x<=1e8) and k (1<=k<=10), k integer a1,a2,..,ak (0<=ai<=9 for i=1..k and ai!=aj for i!=j) is following.

输出描述:

For each test case output your answer y. Your answer should be a positive integer without leading zero and should be no more than 1e18. Every answer that satisfy the conditions descripted above will be accepted.

示例1

输入

复制

3
5 3 1 5 7
21 4 2 5 6 9
10 9 0 1 2 3 4 5 6 7 9

输出

复制

175
2592576
976543210

说明

175 is the multiple of 5 and contains 1,5,7

2592576 is the multiple of 21 and contains 2,5,6,9

976543210 is the multiple of 10 and contains 0,1,2,3,4,5,6,7,9

备注:

Constraint of data

1<=T<=10000

1<=x<=1e8

1<=k<=10

0<=ai<=9,for i=1..k

ai!=aj for i!=j

your answer y should satisfy y <= 1e18

读懂题

题意• 
找一个数(<=1e18)，包含a1,a2,…,ak且是x的倍数。(0<=ai<=9,x<=1e8)
题解• 
令N=102345678900000000
• 1) N%x==0 , ans = N
• 2) N%x!=0, ans = N+(x-N%x)
• 那么显然对于x<=1e8, ans包含0到9且ans是x的倍

 

#include <stdio.h>

long long n = 123456789000000000LL;


int main()
{
    int T,k,t,r;
    char buf[111];
    scanf("%d",&T);
    getchar();
    while(fgets(buf,100,stdin))
    {
        sscanf(buf,"%d",&k);
        r = n % k;
        t = k - r;
        printf("%lld\n",n+t);
    }
    return 0;
}

 

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cfloat>
#include <climits>
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <bitset>
using namespace std;

#define LL long long
const int maxn = 20;
int T;
LL x, k, ans;
int num[maxn];

int main()
{
#ifdef Dmaxiya
    freopen("test.txt", "r", stdin);
#endif // Dmaxiya
    ios::sync_with_stdio(false);

    scanf("%d", &T);
    while(T--)
    {
        ans = 987654321000000000LL;
        scanf("%lld%lld", &x, &k);
        for(int i = 0; i < k; ++i)
        {
            scanf("%d", &num[i]);
        }
        printf("%lld\n", ans + (x - ans % x));
    }

    return 0;
}