Codeforces Round #324 (Div. 2) A. Olesya and Rodion 构造数字思维题-白红宇

Codeforces Round #324 (Div. 2) A. Olesya and Rodion 构造数字思维题

阅读量：4112 次

发布时间：2019-05-25

本文共 2096 字，大约阅读时间需要 6 分钟。

Olesya and Rodion

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.

Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print ?-?1.

Input

The single line contains two numbers, n and t (1?≤?n?≤?100, 2?≤?t?≤?10) — the length of the number and the number it should be divisible by.

Output

Print one such positive number without leading zeroes, — the answer to the problem, or ?-?1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.

Sample test(s)

input

3 2

        output 

题意：要求构造一个n位数字，使得能被t整除

错因分析：做题的思维僵化，太过“暴力”

AC代码：

#include 
       
         #include
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include
               #include 
               
                 #include 
                
                  using namespace std; #define MM(a) memset(a,0,sizeof(a)) typedef long long ll; typedef unsigned long long ULL; const int mod = 1000000007; const double eps = 1e-10; const int inf = 0x3f3f3f3f; int main() { int n, t; while (~scanf("%d %d", &n, &t)) { if (t != 10) for (int i = 1; i <= n; i++) cout << t; else if (n == 1) cout << "-1" ; else { cout << "1"; for (int i = 1; i <= n - 1; i++) cout << "0"; } cout << endl; } return 0; } Wa代码

#include 
       
         #include
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               #include
               #include 
               
                 #include 
                
                  using namespace std; #define MM(a) memset(a,0,sizeof(a)) typedef long long ll; typedef unsigned long long ULL; const int mod = 1000000007; const double eps = 1e-10; const int inf = 0x3f3f3f3f; int main() { int n,t; while (~scanf("%d %d",&n,&t)) { double l = 1en;//10^n，，会编译错误 while (l%t != 0) l++; printf("%.0f\n", l); } return 0; }

转载地址：http://cvgsi.baihongyu.com/

你可能感兴趣的文章