洛谷 P2309 loidc 卖卖萌

https://www.luogu.org/problemnew/show/P2309

思路
求一下前缀和，如有Si-Sj>0，即Si>Sj
则j+1到i的子段和>0
ans = 正序对个数+0到i的子段和中>0的个数即前缀和>0的个数

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
ll n,ans;
ll num[5000010],sum[5000010],tmp[5000010];
void merge_s(ll l,ll r)
{
    if(l==r)
    return;
    ll mid=(l+r)>>1;
    merge_s(l,mid);
    merge_s(mid+1,r);
    ll i=l,j=mid+1,k=l;
    while(i<=mid&&j<=r)
    {
        if(num[i]>num[j])
        {
            tmp[k++]=num[j++];
            ans+=mid-i+1;
        }
        else
        tmp[k++]=num[i++];
    }
    while(i<=mid)
    tmp[k++]=num[i++];
    while(j<=r)
    tmp[k++]=num[j++];
    for(ll i=l;i<=r;i++)
    num[i]=tmp[i]; 
} 
int main()
{
    scanf("%lld",&n);
    for(ll i=1;i<=n;i++)
    {
        scanf("%lld",&num[i]);
        sum[i]=sum[i-1]+num[i];
    }
    for(ll i=1;i<=n;i++)
    {
        num[i]=sum[n-i+1];//将前缀和数列倒过来求逆序对 
        if(num[i]>0)//从0到n-i+1位置的子串(前缀)和>0 
        ans++;
    }
    merge_s(1,n);
    printf("%lld",ans);
    return 0;
}