归并排序模板

C++模板

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

    int mid = (l + r)/2;

    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{
            res += mid - i + 1;
            tmp[k++] = q[j++];
        }
    }
    while(i <= mid) tmp[k++] = q[i++];
    while(j <= r) tmp[k++] = q[j++];

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

    return res;
}

C模板

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

    int mid = (l + r)/2;

    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{
            res += mid - i + 1;
            tmp[k++] = q[j++];
        }
    }
    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;
}

题目描述：

给定一个长度为 n 的整数数列，请你计算数列中的逆序对的数量。

逆序对的定义如下：对于数列的第 i 个和第 j 个元素，如果满足 ia[j]，则其为一个逆序对；否则不是。

输入格式
第一行包含整数 n，表示数列的长度。

第二行包含 n 个整数，表示整个数列。

输出格式
输出一个整数，表示逆序对的个数。

数据范围

1≤n≤100000，
数列中的元素的取值范围 [1,10^9]。

输入样例：

6
2 3 4 5 6 1

输出样例：

5

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#include<iostream>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;
int n;
int q[N], tmp[N];

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

    int mid = (l + r)/2;

    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{
            res += mid - i + 1;
            tmp[k++] = q[j++];
        }
    }
    while(i <= mid) tmp[k++] = q[i++];
    while(j <= r) tmp[k++] = q[j++];

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

    return res;
}

int main(){
    cin >> n;

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

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

    return 0;
}


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#include<stdio.h>

typedef long long LL;

const int N = 1e5 + 10;
int n;
int q[100010], tmp[100010];

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

    int mid = (l + r)/2;

    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{
            res += mid - i + 1;
            tmp[k++] = q[j++];
        }
    }
    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(){
	int i;
    scanf("%d",&n);

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

    printf("%lld", merge_sort(0, n - 1));  //long long型输出要用 %lld

    return 0;
}