链接：https://www.nowcoder.com/acm/contest/148/D
来源：牛客网
题目描述
Prefix Sum is a useful trick in data structure problems.
For example, given an array A of length n and m queries. Each query gives an interval [l,r] and you need to calculate . How to solve this problem in O(n+m)? We can calculate the prefix sum array B in which Bi is equal to
. And for each query, the answer is Br-Bl-1.
Since Rikka is interested in this powerful trick, she sets a simple task about Prefix Sum for you:
Given two integers n,m, Rikka constructs an array A of length n which is initialized by Ai = 0. And then she makes m operations on it.
There are three types of operations:
1. 1 L R w, for each index i ∈ [L,R], change Ai to Ai + w.
2. 2, change A to its prefix sum array. i.e., let A' be a back-up of A, for each i ∈ [1,n], change Ai to 
3. 3 L R, query for the interval sum .
输入描述:
The first line contains a single number t(1≤ t ≤ 3), the number of the testcases.
For each testcase, the first line contains two integers n,m(1 ≤ n,m ≤ 105).
And then m lines follow, each line describes an operation(1 ≤ L ≤ R≤ n, 0 ≤ w ≤ 109).
The input guarantees that for each testcase, there are at most 500 operations of type 3.输出描述:
For each query, output a single line with a single integer, the answer modulo 998244353.示例1
输入
1
100000 7
1 1 3 1
2
3 2333 6666
2
3 2333 6666
2
3 2333 6666
输出
13002
58489497
12043005题意：长度为n的数组，初始值都为0，操作1 将L R区间的数增加w，操作2将数组求一次前缀和，操作3 输出L R区间的数之和
思路：一开始考虑线段树，因为输出最多500次，所以考虑只在输出的时候求前缀和，前面的求前缀和记录下来，但是由于还有区间增加操作，所以一直没能处理好。赛后看别人代码是组合数， 一脸懵逼，然后看了看自己的草稿，看了看杨辉三角的组合数的图，才醒悟。
首先，图一是杨辉三角，图二是多次求前缀和的过程，前缀和数组当前位=原数组当前位+前缀和数组前一位，如果这样不够清楚的话，可以将你的头往左边旋转45°来看。
所以，当遇到操作1时，记录l r w 还有当前的2操作个数，当遇到2操作时，2操作数增加，遇到3操作时，可以通过组合数求得答案。
但是呢，就算你知道了要用组合数求，后面思考还是很麻烦的，我没怎么想明白，这篇博客写的挺好的，可以参照2018牛客网暑假ACM多校训练赛（第十场）D Rikka with Prefix Sum 组合数学，主要是取反前缀这边比较难理解。
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
const ll mod=998244353;
const int maxn=200005;
ll mo(ll a,ll pp){
if(a>=0&&a<pp)return a;
a%=pp;
if(a<0)a+=pp;
return a;
}
ll powmod(ll a,ll b,ll pp){
ll ans=1;
for(;b;b>>=1,a=mo(a*a,pp)){
if(b&1)ans=mo(ans*a,pp);
}
return ans;
}
ll b[maxn],invf[maxn],inv[maxn];
ll C(int n,int m){
if(n<m)return 0;
if(m==n||m==0)return 1;
return b[n]*invf[n-m]%mod*invf[m]%mod;
}
void init(){
b[0]=1;//累乘
for(int i=1;i<maxn;i++)b[i]=b[i-1]*i%mod;
inv[1]=1;//逆元
for(int i=2;i<maxn;i++)inv[i]=((mod-mod/i)*inv[mod%i])%mod;
invf[0]=1;//累乘逆元
for(int i=1;i<maxn;i++)invf[i]=(invf[i-1]*inv[i])%mod;
}
ll inv1(ll b){
return powmod(b,mod-2,mod);
}
ll inv2(ll a){
if(a==1)return 1;
return inv2(mod%a)*inv2(mod-mod/a)%mod;
}
struct node{
int l,r,w,x;
}a[maxn];
int cnt;
int qsum;
ll solve(int x){
int k=qsum+1;
ll ans=0;
for(int i=0;i<cnt;i++){
if(a[i].l<x){
ans=(ans+C(k-a[i].x+1+x-a[i].l-1,k-a[i].x+1-1)%mod*a[i].w%mod+mod)%mod;
}
else if(a[i].l==x){
ans=(ans+a[i].w+mod)%mod;
}
}
return ans;
}
int main(){
int t;
scanf("%d",&t);
init();
while(t--){
int n,m;
scanf("%d%d",&n,&m);
qsum=-1;
int l,r,w;
cnt=0;
while(m--){
int op;
scanf("%d",&op);
if(op==2){
qsum++;
}
else if(op==1){
scanf("%d%d%d",&l,&r,&w);
a[cnt].l=l;
a[cnt].w=w;
a[cnt].x=qsum;
cnt++;
a[cnt].l=r+1;
a[cnt].w=(mod-w)%mod;//直接写-w就TLE了
a[cnt].x=qsum;
cnt++;
}
else{
scanf("%d%d",&l,&r);
printf("%lld\n",(solve(r)-solve(l-1)+mod)%mod);
}
}
}
return 0;
}