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  • HDU 2132 An easy problem

    http://acm.hdu.edu.cn/showproblem.php?pid=2132

    Problem Description
    We once did a lot of recursional problem . I think some of them is easy for you and some if hard for you.
    Now there is a very easy problem . I think you can AC it.
      We can define sum(n) as follow:
      if i can be divided exactly by 3 sum(i) = sum(i-1) + i*i*i;else sum(i) = sum(i-1) + i;
      Is it very easy ? Please begin to program to AC it..-_-
     
    Input
      The input file contains multilple cases.
      Every cases contain only ont line, every line contains a integer n (n<=100000).
      when n is a negative indicate the end of file.
     
    Output
      output the result sum(n).
     
    Sample Input
    1
    2
    3
    -1
     
    Sample Output
    1
    3
    30

     代码:

    #include <bits/stdc++.h>
    using namespace std;
    
    const int maxn = 1e5 + 10;
    long long sum[maxn];
    
    int main() {
        sum[0] = 0;
        for(long long i = 1; i < maxn; i ++) {
            if(i % 3 == 0)
                sum[i] = sum[i - 1] + i * i * i;
            else
                sum[i] = sum[i - 1] + i;
        }
        int x;
        while(~scanf("%d", &x)) {
            if(x < 0)
                break;
            else
                printf("%lld
    ", sum[x]);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9451327.html
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