归并:将序列转化为多个小区间处理后再归并

二路归并

$1.$归并排序

int n;
int q[], tmp[];

void merge_sort(int q[], int l, int r)
{
    if (l >= r) return;

    int mid = l + r >> 1;

    merge_sort(q, l, mid);
    merge_sort(q, mid + 1, r);

    int k = 0, i = l, j = mid + 1;

    while (i <= mid && j <= r) // 归并两个数组为一个tmp一个数组
        if (q[i] <= q[j]) tmp[k ++] = q[i ++];
        else tmp[k ++] = q[j ++];

    while (i <= mid) tmp[k ++] = q[i ++]; // 如果q[l ~ mid]还没归并完，剩下的全部归并至tmp中
    while (j <= r) tmp[k ++] = q[j ++];   // 如果q[mid ~ r]还没归并完，剩下的全部归并至tmp中

    for (i = l, j = 0; i <= r; i ++, j ++) q[i] = tmp[j];
}

$2.$求逆序对数量

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

using namespace std;

typedef long long LL;

const int N = 500050;

int n;
int q[N], tmp[N];

LL merge_sort(int l, int r)
{
    if (l >= r) return 0;

    int mid = l + r >> 1;
    LL res = merge_sort(l, mid) + merge_sort(mid + 1, r);

    int i = l, j = mid + 1, k = 0;
    while (i <= mid && j <= r)
        if (q[i] <= q[j]) tmp[k ++] = q[i ++];
        else
        {
            tmp[k ++] = q[j ++];
            res += mid - i + 1;
        }
    while (i <= mid) tmp[k ++] = q[i ++];
    while (j <= r) tmp[k ++] = q[j ++];

    for (i = l, j = 0; i <= r; i ++, j ++) q[i] = tmp[j];
    return res;
}

int main()
{
    scanf ("%d", &n);

    for (int i = 0; i < n; i ++) scanf ("%d", &q[i]);

    printf ("%lld", merge_sort(0, n - 1));
    return 0;
}

多路归并

1. 技能升级

思路:先找到所有等差数列合并以后的第$M$小的数$x$(注意$ \geq x$的个数可能$\geq M$),再求出前M大的数的和即可

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

using namespace std;

typedef long long LL;

const int N = 100010;

int n, m;
int a[N], b[N];

bool check(int x)
{
    LL res = 0;

    for (int i = 1; i <= n; i ++)
        if (a[i] >= x)
            res += (a[i] - x) / b[i] + 1;

    return res >= m;
}

int main()
{
    scanf ("%d %d", &n, &m);

    for (int i = 1; i <= n; i ++) scanf ("%d %d", &a[i], &b[i]);

    int l = 0, r = 1e6;
    while (l < r)
    {
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }

    LL res = 0, cnt = 0;

    for (int i = 1; i <= n; i ++)
        if (a[i] >= r)
        {
            int c = (a[i] - r) / b[i] + 1; // 项数
            cnt += c; // 总项数
            int end = a[i] - (c - 1) * b[i]; // 大于等于r的最后一个数
            res += (LL) (a[i] + end) * c / 2; // 高斯求和
        }

    printf ("%lld", res - (cnt - m) * r);
    return 0;
}