SGU105 Div 3

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最新推荐文章于 2017-03-05 17:24:12 发布

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分类专栏：数学 SGU 文章标签： div numbers output input

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105. Div 3

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3.

Input

Input contains N (1<=N<=2³¹ - 1).

Output

Write answer to the output.

Sample Input

Sample Output

用d[i]表示第i个数模3后的余数，则d[i+1] = d[i] + (i + 1) % 3，由此可得数组d的周期为3。

#include <cstdio>

int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        int ans = 0;
        if (n % 3 == 2)
            ans++;
        ans += 2 * (n / 3);
        printf("%d\n", ans);
    }
}