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  • HDU1977 Consecutive sum II【数学计算+水题】

    Consecutive sum II

    Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

    Total Submission(s): 3061    Accepted Submission(s): 1533
    Problem Description
    Consecutive sum come again. Are you ready? Go ~~
    1    = 0 + 1
    2+3+4    = 1 + 8
    5+6+7+8+9  = 8 + 27

    You can see the consecutive sum can be representing like that. The nth line will have 2*n+1 consecutive numbers on the left, the first number on the right equal with the second number in last line, and the sum of left numbers equal with two number’s sum on the right.
    Your task is that tell me the right numbers in the nth line.
    Input
    The first integer is T, and T lines will follow.
    Each line will contain an integer N (0 <= N <= 2100000).
    Output
    For each case, output the right numbers in the Nth line.
    All answer in the range of signed 64-bits integer.
    Sample Input
    3 0 1 2
    Sample Output
    0 1 1 8 8 27
    Author
    Wiskey
    Source

    问题链接HDU1977 Consecutive sum II

    题意简述:参见上文。

    问题分析(略)

    程序说明:(略)

    题记说好的找规律,要找出其规律也许没那么容易,知道了也容易了。


    AC的C语言程序如下:

    /* HDU1977 Consecutive sum II */
    
    #include <stdio.h>
    
    int main(void)
    {
        int t, n;
    
        scanf("%d", &t);
        while(t--) {
            scanf("%d", &n);
    
            printf("%lld %lld
    ", (long long) n * n * n, (long long) (n + 1) * (n + 1) * (n + 1));
        }
    
        return 0;
    }



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  • 原文地址:https://www.cnblogs.com/tigerisland/p/7563626.html
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