本题链接：CSP 202112-2 序列查询新解

本博客给出本题截图：

C++

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

typedef long long LL;

const int N = 100010;

int a[N], b[2 * N];
LL res, cnt;

int main()
{
    int n, m;
    cin >> n >> m;

    a[n + 1] = m;
    for (int i = 1; i <= n; i ++ )
        cin >> a[i];
    
    int t = m / (n + 1);
    for (cnt = 1; cnt * t < m; cnt ++ )
        b[cnt] = cnt * t;
    b[cnt ++ ] = m;

    int i = 0, j = 0;
    while (i < n + 1 || j < cnt - 1)
    {
        if (a[i] < b[j])
        {
            res += (min(a[i + 1], b[j]) - a[i]) * abs(j - i - 1);
            i ++;
        }
        else if (a[i] == b[j])
        {
            res += (min(a[i + 1], b[j + 1]) - a[i]) * abs(i - j);
            i ++;
            j ++;
        }
        else
        {
            res += (min(b[j + 1], a[i]) - b[j]) * abs(i - j - 1);
            j ++;
        }
    }

    cout << res << endl;

    return 0;
}

  
 

  
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总结

先放代码，后续补题解，期末有点忙…

文章来源: chen-ac.blog.csdn.net，作者：辰chen，版权归原作者所有，如需转载，请联系作者。

原文链接：chen-ac.blog.csdn.net/article/details/122379139