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  • A hard puzzle ACM HDU1097

    A hard puzzle

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 13437    Accepted Submission(s): 4677


    Problem Description
    lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.
    this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.
     

    Input
    There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)
     

    Output
    For each test case, you should output the a^b's last digit number.
     

    Sample Input
    7 66 8 800
     

    Sample Output
    9 6
     

    Author
    eddy
     

    Recommend
    JGShining
     
    #include<stdio.h>
    int main()
    {
    int a,b;
    int t;
    int res;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
    t
    =a%10;
    res
    =1;
    while(b)
    {
    if(b%2!=0)
    {
    res
    *=t;
    res
    %=10;
    }
    t
    =t*t;
    t
    %=10;
    b
    /=2;
    }
    printf(
    "%d\n",res);
    }
    }

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  • 原文地址:https://www.cnblogs.com/kuangbin/p/2123488.html
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