luogu2221 [HAOI2012]高速公路

和sdoi的相关分析很像qwq，推柿子然后线段树搞搞

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
int n, m, uu, vv, ww;
ll que[3];
char ss[15];
struct SGT{
	ll sum[400005][3], tag[400005];
	//sum[0]: \sum w_i i, sum[1]: \sum w_i i^2, sum[2]: \sum w_i
	void pushDown(int o, int l, int r, int lson, int rson, int mid){
		tag[lson] += tag[o];
		tag[rson] += tag[o];
		sum[lson][0] += (ll)tag[o]*(mid-l+1)*(l+mid)/2;
		sum[lson][1] += (ll)tag[o]*((ll)mid*(mid+1)*(2*mid+1)/6-(ll)(l-1)*l*(2*l-1)/6);
		sum[lson][2] += (ll)tag[o]*(mid-l+1);
		sum[rson][0] += (ll)tag[o]*(r-mid)*(mid+1+r)/2;
		sum[rson][1] += (ll)tag[o]*((ll)r*(r+1)*(2*r+1)/6-(ll)mid*(mid+1)*(2*mid+1)/6);
		sum[rson][2] += (ll)tag[o]*(r-mid);
		tag[o] = 0;
	}
	void update(int o, int l, int r, int x, int y, int k){
		if(l>=x && r<=y){
			sum[o][0] += (ll)k*(r-l+1)*(l+r)/2;
			sum[o][1] += (ll)k*((ll)r*(r+1)*(2*r+1)/6-(ll)(l-1)*l*(2*l-1)/6);
			sum[o][2] += (ll)k*(r-l+1);
			tag[o] += k;
		}
		else{
			int mid=(l+r)>>1;
			int lson=o<<1;
			int rson=lson|1;
			if(tag[o])	pushDown(o, l, r, lson, rson, mid);
			if(x<=mid)	update(lson, l, mid, x, y, k);
			if(mid<y)	update(rson, mid+1, r, x, y, k);
			for(int i=0; i<3; i++)
				sum[o][i] = sum[lson][i] + sum[rson][i];
		}
	}
	void query(int o, int l, int r, int x, int y){
		if(l>=x && r<=y)
			for(int i=0; i<3; i++)
				que[i] += sum[o][i];
		else{
			int mid=(l+r)>>1;
			int lson=o<<1;
			int rson=lson|1;
			if(tag[o])	pushDown(o, l, r, lson, rson, mid);
			if(x<=mid)	query(lson, l, mid, x, y);
			if(mid<y)	query(rson, mid+1, r, x, y);
		}
	}
}sgt;
ll gcd(ll a, ll b){
	return !b?a:gcd(b, a%b);
}
int main(){
	cin>>n>>m;
	while(m--){
		scanf("%s", ss);
		if(ss[0]=='C'){
			scanf("%d %d %d", &uu, &vv, &ww);
			vv--;
			sgt.update(1, 1, n, uu, vv, ww);
			sgt.query(1, 1, n, 1, 1);
		}
		else{
			scanf("%d %d", &uu, &vv);
			que[0] = que[1] = que[2] = 0;
			ll fm=(ll)(vv-uu+1)*(vv-uu)/2;
			vv--;
			sgt.query(1, 1, n, uu, vv);
			ll ans=(ll)(uu+vv)*que[0]-que[1]+(ll)(vv+1)*(1-uu)*que[2];
			ll f=gcd(ans, fm);
			printf("%lld/%lld\n", ans/f, fm/f);
		}
	}

	return 0;
}