【算法】-前缀和之前缀和的前缀和

思想

计算前缀和
python计算前缀和有两种方法，一种是直接计算，一种是调用库函数

       # 方法1：计算前缀和
        n = len(a)
        s = [0] * (n+1) # 第一个存储的0，s[i] = a0+...ai-1
        for i, v in enumerate(a):
            s[i+1] = s[i]+v

        for i, v in enumerate(s):
            ss[i+1] = s[i] + v

        # 方法2：调用库函数
        s = accumulate(a, initial=0)
        # 计算前缀和的和
        ss = list(accumulate(s,initial=0))

使用c++进行计算前缀和

        int n = a.size()
        vector<int> pre_sum(n+1,0),pre_pre_sum(n+1,0);
        rep(i,0,n)pre[i+1] = pre[i] + a[i];
        rep(i,0,n)pre_pre_sum = pre_pre_sum[i] + pre[i];

模板
#include<bits/stdc++.h>
#define D(...) fprintf(stderr,__VA_ARGS__)
#define DD(...) D(#__VA_ARGS__ "="),debug_helper::debug(__VA_ARGS__),D("\n")
#include"/home/xay5421/debug.hpp"
#else
#define D(...) ((void)0)
#define DD(...) ((void)0)
#endif
#define pb push_back
#define eb emplace_back
#define SZ(x) ((int)(x).size())
#define each(x,v) for(auto&x:v)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;
using LL=long long;
using ULL=unsigned long long;
const int P=1e9+7;
int ad(int k1,int k2){return k1+=k2-P,k1+=k1>>31&P;}
int su(int k1,int k2){return k1-=k2,k1+=k1>>31&P;}
int mu(int k1,int k2){return 1LL*k1*k2%P;}
void uad(int&k1,int k2){k1+=k2-P,k1+=k1>>31&P;}
void usu(int&k1,int k2){k1-=k2,k1+=k1>>31&P;}
template<class... T>int ad(int k1,T... k2){return ad(k1,ad(k2...));}
template<class... T>int su(int k1,T... k2){return su(k1,ad(k2...));}
template<class... T>int mu(int k1,T... k2){return mu(k1,mu(k2...));}
template<class... T>void uad(int&k1,T... k2){return uad(k1,ad(k2...));}
template<class... T>void usu(int&k1,T... k2){return usu(k1,ad(k2...));}
int po(int k1,int k2){
	int k3=1;
	for(;k2;k2>>=1,k1=mu(k1,k1))if(k2&1)k3=mu(k3,k1);
	return k3;
}
const int N=100005;
int n;
int a[N],st[N],top,lc[N],rc[N];
int s1[N],s2[N],s3[N];
int ans;
int calc1(int l,int r){
	return su(s1[r],s1[l-1],mu(l-1,su(s3[r],s3[l-1])));
}
int calc2(int l,int r){
	return ad(su(s2[r],s2[l-1]),mu(r+1,su(s3[r],s3[l-1])));
}
void dfs(int u,int l,int r){
	if(l>r)return;
	uad(ans,mu(a[u],ad(mu(calc1(l,u),r-u+1),mu(calc2(u+1,r),u-l+1))));
	dfs(lc[u],l,u-1);
	dfs(rc[u],u+1,r);
}