博客
关于我
CSUOJ 1018 Avatar
阅读量:428 次
发布时间:2019-03-06

本文共 1477 字,大约阅读时间需要 4 分钟。

Description

In the planet Pandora, Jake found an old encryption algorithm. The plaintext, key and ciphertext are all four decimal numbers and all greater than 0. The way to get the ciphertext from the plaintext and the key is very simple: the ciphertext is the last four digits of the product of the plaintext and the key (for example: the plaintext is 2010 and the key is 4024, and then the product is 8088240. So the ciphertext is 8240).

Note that the plaintext and the key don’t have leading 0, while the ciphertext might have. Now given the plaintext and the key, you are asked to figure out the ciphertext for Jake quickly.

Input

The first line is an integer T, which presents the number of test cases. Then there are T lines, each line has two integers, the first integer is the plaintext and the second is the key.

Output

For each test case, output the ciphertext.

Sample Input

22010 40241234 1111

Sample Output

82400974

Hint


两数相乘,取模10000就可以了
#include<stdio.h>#include<string>#include<string.h>#include<algorithm>#include<iostream>typedef long long ll;using namespace std;ll x, y;int main(){	int t;	while (~scanf("%d", &t))	{		while (t--)		{			scanf("%lld%lld", &x, &y);			ll num = x*y;			num = num % 10000;			printf("%.4lld\n", num);		}	}	return 0;}/**********************************************************************	Problem: 1018	User: leo6033	Language: C++	Result: AC	Time:0 ms	Memory:2024 kb**********************************************************************/

转载地址:http://tzwuz.baihongyu.com/

你可能感兴趣的文章
mysql之分组查询GROUP BY,HAVING
查看>>
mysql之分页查询
查看>>
Mysql之备份与恢复
查看>>
mysql之子查询
查看>>
MySQL之字符串函数
查看>>
mysql之常见函数
查看>>
Mysql之性能优化--索引的使用
查看>>
mysql之旅【第一篇】
查看>>
Mysql之索引选择及优化
查看>>
mysql之联合查询UNION
查看>>
mysql之连接查询,多表连接
查看>>
mysql乱码
查看>>
Mysql事务。开启事务、脏读、不可重复读、幻读、隔离级别
查看>>
MySQL事务与锁详解
查看>>
MySQL事务原理以及MVCC详解
查看>>
MySQL事务及其特性与锁机制
查看>>
mysql事务理解
查看>>
MySQL事务详解结合MVCC机制的理解
查看>>
MySQL事务隔离级别:读未提交、读已提交、可重复读和串行
查看>>
MySQL事务隔离级别:读未提交、读已提交、可重复读和串行
查看>>