CodeForces 584A
2017-03-05 01:12
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C -C
Crawling in process...Crawling failed
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
SubmitStatusPractice
CodeForces 584A
Description
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible byt. 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 andt (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 Input
Input
Output
思路:水题,一想便知;
Crawling in process...Crawling failed
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
SubmitStatusPractice
CodeForces 584A
Description
Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible byt. 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 andt (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 Input
Input
3 2
Output
712
思路:水题,一想便知;
#include<cstdio> int main() { int n,t,i; scanf("%d %d",&n,&t); if(n==1 && t==10) printf("-1\n"); else if(t==10) { printf("1"); for(i=0;i<n-1;i++) printf("0"); printf("\n"); } else { for(i=0;i<n;i++) { printf("%d",t); } printf("\n"); } return 0; }
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