CodeForces 248B Chilly Willy
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Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
A single input line contains a single integer n (1 ≤ n ≤ 105).
Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
1
-1
5
10080
题意:n位数,最小的,是2 3 5 7的倍数。。。
思路:先打表找规律
#include<iostream>#include<stdio.h>#include<string.h>#include<cmath>using namespace std;#define mod 1000000007#define ll long longint main(){ int n; /*while(cin>>n) { ll i=pow(10,n-1); for(;i<i*10;i++) { if(i%210==0) break; } cout<<i<<endl; }*/ string map[]={"","050","080","170","020","200","110"}; while(cin>>n) { if(n<3) cout<<-1<<endl; else if(n==3) cout<<210<<endl; else { cout<<1; for(int i=1;i<n-3;i++) cout<<0; n=(n+3)%6; if(n==0) n=6; cout<<map[n]<<endl; } } return 0;}
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