鏈接：https://ac.nowcoder.com/acm/contest/887/D
來源：牛客網(wǎng)

題目描述

I have a very simple problem for you.?Given a positive integeter?n?(1≤n≤1000000)n\ (1 \leq n \leq 1000000)n?(1≤n≤1000000) and a prime number?p?(2≤p≤1000000)p\ (2 \leq p \leq 1000000)p?(2≤p≤1000000), your job is to output a positive number which is divisible by and has exactly digits. Please output "T_T" if you can not find such number.

輸入描述:

The first line of the input file contains two integers?n?(1≤n≤1000000)n\ (1 \leq n \leq 1000000)n?(1≤n≤1000000) and p?(2≤p≤1000000)p\ (2 \leq p \leq 1000000)p?(2≤p≤1000000).?is a prime number.

輸出描述:

Output one number (without leading zeros) or "T_T" 示例1

輸入

復(fù)制 2 5

輸出

復(fù)制 10 示例2

輸入

復(fù)制 1 11

輸出

復(fù)制 T_T 示例3

輸入

復(fù)制 5 2

輸出

復(fù)制 10000
思路：一開始讀題，理解成了最小的那個(gè)能整除p的n位數(shù)，原來忽略了一個(gè)positive...心累，算了，菜真是沒辦法
#include <cstdio> #include <iostream> #include <cmath> #include <string> #include <cstring> #include <algorithm> #include <queue> #include <vector> #include <map> using namespace std;#define ll long long #define eps 1e-9const int inf = 0x3f3f3f3f; const int mod = 1e9+7;int n, p, ans, q, len;int main() {scanf("%d%d", &n, &p);q = p;len = 0;while(q){len++;q /= 10;}if(len > n)printf("T_T\n");else{printf("%d", p);n = n - len;while(n){printf("0");n--;}printf("\n");}return 0; }

?

轉(zhuǎn)載于:https://www.cnblogs.com/RootVount/p/11346340.html

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