CSU1018: Avatar

1018: Avatar

Submit Page   Summary   Time Limit: 1 Sec     Memory Limit: 128 Mb     Submitted: 841     Solved: 557

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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

2
2010 4024
1234 1111

Sample Output

8240
0974

题意：水题，给两个数求相乘后的结果的后四位，直接乘，然后把数记录到数组中，然后输出即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=;
int T;
int a,b;
int ans[maxn];
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d",&a,&b);
        int temp=a*b;
        for(int i=;i<;i++)
        {
            ans[i]=temp%;
            temp/=;
        }
        for(int i=;i>=;i--)
            printf("%d",ans[i]);
        printf("\n");
    }
    return ;
}

/**********************************************************************
    Problem: 1018
    User: therang
    Language: C++
    Result: AC
    Time:0 ms
    Memory:2024 kb
**********************************************************************/